A balloon at an amusement park has 1,500 g of gas. By what amount will the mass of the balloon decrease if all the gas inside it escapes?

- 750 g

- 1,500 g

-2,000 g

- 3,000 g

SOMEONE PLEASE HELP ME IT WILL BE SOOOO APPRECAITED!

Answers

Answer 1
The answer is best explained in the diagram below.

Let m =  the mass of the balloon material (g).

At state 1, the total mass of the balloon is the mass of the balloon material and the mass of gas.
The total mass in state 1 is
m₁ =(m + 1500) g

At state 2, 1500 g of gas has escaped. Only the mass of the balloon material remains.
The total mass in state 2 is
m₂ = m g.

The decrease in mass of the balloon is
m₁ - m₂ = (m + 1500) - m = 1500 g

Answer: 1,500 g


A Balloon At An Amusement Park Has 1,500 G Of Gas. By What Amount Will The Mass Of The Balloon Decrease
Answer 2

Answer:

Answer 1,500 g

because if it has 1,500 g and it loses ALL of the gas...

then it will lose 1,500 g

Hope this Helps!


Related Questions

Person is lifting a 250 N dumbbell. The weight is 30 cm from the pivot point of the elbow. What force must be exerted five from the elbow to lift the weight? Assume everything is perpendicular.

Answers

Refer to the diagram shown below.

The force, F, is applied at 5 cm from the elbow.

For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N

Answer: 1500 N

Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel is y=2.0×1011n/m2. how far (δl) would such a string stretch under a tension of 1500 n? express your answer in millimeters using two significant figures.

Answers

L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ =  the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer:  15 mm

The steel guitar string stretches by 15 mm under a tension of 1500 N.

To determine how far a steel guitar string stretches under a tension force, we use principles from materials science, specifically Young's modulus and Hooke's law.

Identify the given values:

Initial length, l = 1.00 m

Cross-sectional area, a = 0.500 mm² = 0.500 × 10⁻⁶ m²

Young's modulus, Y = 2.0 × 1011 N/m²

Tension force, F = 1500 N

Use the formula for elongation:

δl = (F × l) / (Y × a)

Substitute the given values into the formula:

[tex]\delta l = \frac{1500 \, \text{N} \times 1.00 \, \text{m}}{2.0 \times 10^{11} \, \text{N/m}^2 \times 0.500 \times 10^{-6} \, \text{m}^2} \\[/tex]

Calculate the result:

[tex]\delta l = \frac{1500 \times 1.00}{2.0 \times 10^{11} \times 0.500 \times 10^{-6}} \\[/tex]

[tex]\delta l = \frac{1500}{1.0 \times 10^5} = 0.015 \, \text{m} = 15 \, \text{mm}[/tex]

The string stretches by 15 mm when a tension of 1500 N is applied.

The force required to compress a non-standard spring as a function of displacement from equilibrium x is given by the equation f(x) = ax2 - bx, where a = 65 n/m2, b = 12 n/m, and the positive x direction is in the compression direction of the spring.

Answers

Final answer:

This is a high school Physics question about the force required to compress a non-standard spring, and how to calculate the work done in the process.

Explanation:

The subject of the student's question pertains to

Physics

. The topics discussed are related to forces and the physical properties of springs, specifically the force required to compress a spring and the work done in the process. This can be addressed by Hooke's law, where the restoring force of the spring is directly proportional to its displacement from equilibrium. The function given, f(x) = ax

2

- bx, represents a non-standard spring since the force is not linearly proportional to the displacement but depends on the square of the displacement. Here, 'a' represents the constant relating force to the squared displacement, and 'b' represents inverse proportionality between force and displacement. For a standard spring, the equation f = -kx is used where F is the restoring force, x is the displacement and k is the spring constant. In this non-standard case, we can integrate the function f(x) from 0 to the point of desired compression to determine the work done by the spring force, using the principle that work done is the integral of force over displacement. This application of mechanics and understanding of forces makes this a

high school Physics question

.

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Final answer:

The question relates to the physics of springs and work done during their compression or extension. The force required to compress or extend a non-standard spring is given by the expression f(x) = ax2 - bx. The work done and potential energy stored depend on the square of the displacement from equilibrium.

Explanation:

The subject matter pertains to the physics of spring forces, specifically the equation for the force required to compress a non-standard spring, f(x) = ax2 - bx. Here, 'a' and 'b' are constants with given values, and 'x' represents the displacement from equilibrium. When a spring is compressed, it exerts a restoring force in the opposite direction. To calculate the work done by this force, the displacement plays an important role as the equation f(x) = ax2 - bx suggests. This equation also applies to extensions, with positive 'x' signifying compression (stretch) and negative 'x' indicating extension.

For instance, if the displacement 'x' is +6 cm (meaning the spring is compressed by 6 cm), we can calculate the work done by substituting this value for 'x' in the equation. The work done also depends on the square of the displacement. Hence, a greater displacement results in more work done by the spring force, and thus more potential energy stored in the spring.

Using Hooke's law, which states the force exerted by a spring is directly proportional to the displacement from its equilibrium position, we can compare the characteristics of a non-standard spring relative to a standard one. This knowledge is pivotal to understanding the behavior of springs in various real-world applications.

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When an object falls, it trades gravitational potential energy for kinetic energy, accelerating toward the ground. calculate the potential energy change an object of mass m would experience falling from a height h. if there were no air friction, so that all of the energy was converted to kinetic energy, what would its final velocity be in terms of g and h?

Answers

the speed of the energy decreases

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker is slipping with a velocity of 8.6 m/s north and the halfback is sliding with a velocity of 7.4 m/s east. What is the magnitude of the velocity at which the two players move together immediately after the collision

Answers

Momentum 
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude 

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v      [solve for v]
v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
Final answer:

The two football players move together at a velocity of 5.99 m/s immediately after the collision. This value was obtained by conserving the momentum of the system: calculating individual momenta before the collision, summing them to obtain total momentum, and then dividing this by the total mass of both players. This is a typical conservation of momentum problem in high school level physics.

Explanation:

The situation you're describing is a perfect example of a conservation of momentum problem in physics. In this problem, the two football players can be seen as a system, and their individual momenta before the tackle add up to yield the total momentum of the system after the tackle, when they're moving together.

Momentum is calculated as mass times velocity, so we first calculate the momentum of each player before the collision. For the linebacker: 120kg× 8.6m/s = 1032 kg×m/s north. For the halfback: 75kg× 7.4m/s = 555 kg×m/s east.

These two momentum vectors form a right triangle, with the hypotenuse representing the  result vector or total momentum of the system after the tackle. We can use the Pythagorean theorem to calculate the magnitude of this vector: sqrt((1032²)+(555²)) = 1169 kg*m/s.

Since the players are moving together after the collision, the mass we use to find the final velocity should be the total mass of both players: 120kg + 75kg = 195 kg. The magnitude of the velocity at which the two players move together immediately after the collision can then be obtained by dividing the total momentum by the total mass: 1169kg×m/s / 195kg = 5.99 m/s.

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Ice sheet movement rates have varied from about 50 to 320 meters per year for the margins of the ice sheet advancing from the hudson bay region during the ice age. if an ice sheet moved from the southern end of hudson bay to the south shore of present-day lake erie, a distance of 1600 kilometers, what would be the maximum amount of time required?

Answers

First let us convert the distance into meters.

distance = 1600 km = 1,600,000 m

 

Then we get the maximum time by dividing the distance with the smallest movement rate possible, that is:

maximum time = 1,600,000 m / (50 m / year)

maximum time = 32,000 years

Final answer:

Using the minimum movement rate of 50 meters per year, the maximum amount of time required for an ice sheet to travel 1600 kilometers from Hudson Bay to Lake Erie is calculated to be 32,000 years.

Explanation:

The question revolves around calculating the maximum amount of time required for an ice sheet to move from the southern end of Hudson Bay to the south shore of present-day Lake Erie, a distance of 1600 kilometers, given varying movement rates.

First, convert kilometers to meters since the movement rates are given in meters per year. The distance is 1,600 kilometers, which equals 1,600,000 meters. Since we are interested in the maximum amount of time required, we will use the minimum speed of 50 meters per year. This will yield the longest possible time it would take for the ice sheet to cover this distance.

To find the time, we use the formula: Time = Distance / Speed. Substituting the given values: Time = 1,600,000 meters / 50 meters per year = 32,000 years.

Therefore, at a movement rate of 50 meters per year, the maximum amount of time required for an ice sheet to travel from Hudson Bay to Lake Erie would be 32,000 years.

2. Why was it important to examine both the color and the streak of your minerals? Think about streak and explain why it’s called a mineral’s “true color”. Answer in at least 2 sentences.

Answers

Some minerals might have similar color, but different streaks. Others might have the same streak color, but different colors

The examination of color and streak are the physical tests that are performed on the minerals to know their exact origin.

Two or more rocks or minerals can have similar color but they are unrelated in their origin and chemical composition. A color test can give false indication that two minerals or two rocks belong to the same type. A streak test can give direct indication of same or different origin as the mineral or rock is pressed against a hard substratum it liberates a powdery color residue which can be used to examine the exact color of the rock or mineral.

Hence, this can be stated that a streak is useful to give true color of the mineral.

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Describe how the energy from stars is release?

Answers

it is produced from the solar energy that surrounds the outer space area 
Final answer:

The energy from stars is released through a process called nuclear fusion, where hydrogen atoms combine to form helium and release a tremendous amount of energy. This energy is in the form of electromagnetic radiation, including visible light, infrared radiation, ultraviolet radiation, and X-rays. Stars release this energy continuously throughout their lifetimes.

Explanation:

The energy from stars is released through a process called nuclear fusion. In the core of a star, hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. This energy is in the form of electromagnetic radiation, which includes visible light, infrared radiation, ultraviolet radiation, and X-rays.

During nuclear fusion, the mass of the combined helium nucleus is slightly less than the mass of the original hydrogen atoms. This mass difference is converted into energy according to Einstein's famous equation, E=mc^2, where E represents energy, m represents mass, and c represents the speed of light.

Stars release this energy continuously throughout their lifetimes, providing the heat and light that sustains life on Earth and allowing astronomers to study distant objects in the universe.

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If you drop an object from a height of 1.9 m, it will hit the ground in 0.62 s. if you throw a baseball horizontally with an initial speed of 34 m/s from the same height, how long will it take the ball to hit the ground?

Answers

That's the cool thing about free fall.  The amount of time it takes to fall remains the same.

In this case, a ball that is simply dropped from rest will fall at the same rate as a ball that had some umph in the horizontal direction.

Assuming the dread pirate roberts never misses, how far from the end of the cannon is the ship that you are trying to hit (neglect dimensions of cannon)? answer in units of m.

Answers

Final answer:

The distance from the cannon to the ship is 32.0 km, but to calculate initial velocity and maximum height of the shell, additional information such as angle of launch or time is required. The Earth's curvature slightly affects the height of the ocean's surface relative to the ship over long distances.

Explanation:

Assuming the dread pirate Roberts never misses, the distance from the end of the cannon to the ship trying to be hit is the maximum distance the cannon shell can travel, which is 32.0 km when neglecting the dimensions of the cannon and air resistance. To calculate the initial velocity of the shell, we would use the kinematic equations for projectile motion. However, the question does not provide angle of launch or time required for the calculation, so further information would be needed to complete part (a).

For part (b), the maximum height reached by the shell cannot be calculated without additional information such as the launch angle or time of flight. For part (c), the curvature of the Earth impacts the level of the ocean's surface in relation to a straight line extending from the ship. Using the Earth's radius (6.37×10³ km), we can apply the principles of geometry to find the drop in height over a distance of 32.0 km, with the assumption that the ocean surface follows the curvature of the Earth.

A force of 80 N is exerted on an object on a frictionless surface for a distance of 4 meters. If the object has a mass of 10 kg, calculate its velocity. v =

Answers

W=ΔKE
F(d)=(1/2)mv^2
v=sqrt((2F(d)/m))
v=sqrt(2(80)(4)/10)
v=sqrt(640/10)
v=8


Answer: The velocity of the object will be 8m/s.

Explanation:

Force exerted on the object = 80 N

Distance displaced by object by the action of force = 4 m

Mass of the object =  10 kg

Velocity gained by the object = v

Kinetic energy of the object =  Work done on the object

[tex]\frac{1}{2}mv^2=Force\times displacement[/tex]

[tex]\frac{1}{2}\times 10\times v^2=80 N\times 4 m[/tex]

[tex]v^2=64 m^2/s^2[/tex]

[tex]v=8 m/s[/tex]

The velocity of the object will be 8m/s.

Scientists have proven that genes play no role in self-esteem. Please select the best answer from the choices provided. T F

Answers

Scientists have proven that genes play no role in self-esteem is a FALSE statement. In fact, most mental health (including self-esteem) has many connections with genetics.

The OXTR gene is said to be related to our self-esteem. Just a fun fact:

Those with 1 or 2 copies of the OXTR with an "A" allele gene are shown to be the more "negative" type of people with lower self-esteems. Those with 2 copies of the OXTR with a "G" allele gene were said to be more optimistic.

Final answer:

The statement that genes do not affect self-esteem is false. Research has found a connection between certain genes and levels of self-esteem, although genes are not the only influencing factors.

Explanation:

The statement that genes play no role in self-esteem is False. It has been scientifically proven that genes play a role in determining an individual's level of self-esteem. Through research studies, it has been observed that there is a connection between specific genes and self-esteem. Such studies involve analyzing the DNA of different individuals and examining the likelihood of these individuals having higher or lower self-esteem, depending on the presence of certain genes.

Genes and Self-Esteem

Research suggests that genes can have an influence on certain personality traits which are associated with self-esteem. For instance, individuals who are naturally more outgoing or socially inclined due to their genetic makeup might have a higher self-esteem than those who are introverted. However, it's important to note that genes are not the only factors that influence self-esteem - environmental factors also play a significant role.

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A yummy glazed doughnut is shown above. 1) where is the center of mass of this fantastic culinary delight? in the center of the hole. somewhere inside the solid part of the doughnut the center of mass is not defined in cases where mass is missing.

Answers

Answer: In the center of the hole.

The center of mass could be calculated by considering the doughnut as consisting of (a) a solid filled-in donut, minus (b) the filled in mass.
Because of symmetry, the center of mass of both masses lies at the same center of the concentric circles.

What is the resistance of a nichrome wire at 0.0 ∘c if its resistance is 200.00 ω at 11.5 ∘c?

Answers

We can use the temperature coefficient of resistance to determine the resistance of the nichrome wire at 0.0 °C. The temperature coefficient of resistance (α) is the amount by which the resistance of a material changes per degree Celsius of temperature change.

Given information:

Resistance of the nichrome wire at 11.5°C = 200.00 Ω

Temperature at which resistance is to be found = 0.0°C

We can use the following formula to find the resistance of the nichrome wire at 0.0°C:

R₂ = R₁ [1 + α (T₂ - T₁)]

Where,

R₁ = Resistance of the wire at temperature T₁

R₂ = Resistance of the wire at temperature T₂

α = Temperature coefficient of resistance

T₁ = Temperature at which R₁ is given

T₂ = Temperature at which R₂ is to be found

Since we are given the resistance of the nichrome wire at 11.5°C, we can take this as R₁ and T₁ as 11.5°C. We also know that the temperature coefficient of resistance for nichrome wire is 0.0004 per °C.

Substituting the given values into the formula, we get:

R₂ = 200.00 Ω [1 + (0.0004/°C) (0.0°C - 11.5°C)]

R₂ = 200.00 Ω [1 - 0.0046]

R₂ = 200.00 Ω (0.9954)

R₂ = 199.08 Ω

Therefore, the resistance of the nichrome wire at 0.0°C is 199.08 Ω.

Nichrome wire resistance at 0.0°C is 199.08 Ω.

Calculate nichrome wire resistance at 0.0°C using its temperature coefficient and resistance at 11.5°C.

The resistance R of a conductor at a temperature T is given by the formula:

[tex]\[ R_T = R_0 \left(1 + \alpha \Delta T\right) \][/tex]

Rearrange the formula to find [tex]\( R_0 \)[/tex] (resistance at 0.0°C):

[tex]\[ R_0 = \frac{R_{11.5}}{1 + \alpha \Delta T} \][/tex]

[tex]\( \Delta T \)[/tex]= 11.5 °C - 0.0°C = 11.5°C

Substitute the values:

[tex]\[ R_0[/tex]=  200.00Ω/(1 + (0.0004°C* 11.5°C))

[tex]\[ R_0[/tex] = 200.00Ω/(1 + 0.0046)

[tex]\[ R_0[/tex] = 200.00Ω/(1.0046)

[tex]\[ R_0[/tex] = 199.08 Ω

A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the meteor's center of mass. What is the gain in Earth's Kinetic Energy?

Answers

Final answer:

The gain in Earth's kinetic energy after being struck by a 50 kg meteorite traveling at 1000 m/s is negligibly small due to Earth's significantly greater mass.

Explanation:

The question asks what the gain in Earth's kinetic energy is when a 50 kg meteorite moving at 1000 m/s strikes it along a line joining their centers of mass. To calculate the gain in kinetic energy, we need to consider the conservation of momentum and understand that Earth's overall movement in space is not significantly altered by this minor collision. Hence, the gain in Earth's kinetic energy is negligibly small because the mass of the meteorite is minuscule compared to the mass of Earth (5.97 × 10^{24} kg). Even at a high velocity, the meteorite's kinetic energy is absorbed, dispersed, and mostly transformed into heat upon impact, rather than causing a noticeable increase in Earth's translational kinetic energy.

In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange one is traveling at a speed of 5 meters per second. The green one is traveling at a speed of 2 meters per second. Which of the forces most affects the motion of the bumper cars after they collide?

Answers

Answer: The force of orange car will affect the motion of the bumper cars after their collision.

Explanation:

Let the mass of both the car be 'm' travelling straight towards each other with time [tex]t_1 and t_2[/tex]

Velocity of orange car =[tex]v_1[/tex] =5m/s

Velocity of green car =[tex]v_2[/tex] =2m/s

Force of orange car =[tex]F_1=ma_1=m\frac{v_1}{t}=m\frac{5m/s}{t_1}[/tex]

[tex]F_1\times t_1=Impulse(I_1)=ma\times 5m/s[/tex]...(1)

Force of green car = [tex]F_2=ma_2=m\frac{v_2}{t}=m\frac{2m/s}{t_2}[/tex]

[tex]F_2\times t_2=Impulse(I_2)=m\times 2m/s[/tex]...(2)

From the above two expression of we see that Impulse is directly proportional to velocity. So, the car with higher velocity will be having impulse. When the orange car bumper car collides with green car the impulse of orange car will be more. With high Impulse the Force on orange car will also be more.

Hence,the force of orange car will affect the motion of the bumper cars after their collision.

Answer:

C. The other car, USA Testprep.

Explanation:

How can genetic engineering help improve crop production?

a.

Genetic engineering can eliminate the threat of disease to crops.

b.

Genetic engineering removes the need of plants for watering.

c.

Genetic engineering can help reduce the effects of pests and weather on crop production.

d.

Genetic engineering can make crops more susceptible to herbicides.

Answers

So the way I'd solve this problem would be to go through each answer and ask myself why it is true/not true. For A, it is true that engineers try to breed disease from their plants. And from just a common sense point of view, wouldn't you want to rid your plants of disease? For B, it doesn't make sense to get rid of watering entirely. That's one of the fundamental things of a plant, besides sunlight and CO2. For C, you can't control the weather, and pests can be stopped using pesticides, which are cheaper than breeding a new plant type. For D, it doesn't make sense to want your plants to grow better from herbicides. So, my answer would be A as it is the best fitting option

Answer:

C). Genetic engineering can help reduce the effects of pests and weather on crop production.

Explanation:

A cue ball has a mass of 0.5 kg. During a game of pool, the cue ball is struck and now has a velocity of 3 . When it strikes a solid ball with a mass of 0.5 kg, the cue ball comes to a complete stop. What is the new velocity of the solid ball? Round your answer to the nearest whole number.

Answers

Answer: 3 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:

[tex]p_i = p_f[/tex]

The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore

[tex]p_i = m_c u_c = (0.5 kg)(3 m/s)=1.5 kg m/s[/tex]

So, the final total momentum will also be

[tex]p_f = 1.5 kg m/s[/tex]

And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:

[tex]p_f = m_s v_s[/tex]

from which we find the velocity of the solid ball

[tex]v_s = \frac{p_f}{m_s}=\frac{1.5 kg m/s}{0.5 kg}=3 m/s[/tex]

3, just did the assignment

A. Calculate the diffraction limit of the human eye, assuming a wide-open pupil so that your eye acts like a lens with diameter 0.8 centimeter, for visible light of 500-nanometer wavelength.
Express your answer using two significant figures.

B. How does this compare to the diffraction limit of a 10-meter telescope?
Express your answer using two significant figures.

C. Now remember that humans have two eyes that are approximately 7 centimeters apart. Estimate the diffraction limit for human vision, assuming that your "optical interferometer" is just as good as one eyeball as large as the separation of two regular eyeballs.
Express your answer using two significant figures.

Answers

Answer:

a) 16 arc seconds

b) 1250

c)1.785 arc seconds

Explanation:

Given data:

lens diameter = 0.8 cm

wavelength 500 nm

a) the diffraction of the eye is given as

[tex]= 2.5\times 10^5 \frac{\lmbda}{D}[/tex] arc seconds

[tex]= 2.5\times 10^5 \frac{5\times 10^{-7}}{8\times 10^{-3}}[/tex] arc seconds

= 16 arc seconds

b) we know that

[tex]\frac{DIffraction\ limit\ of\ eye}{diffraction\ limit\ of\telescope}[/tex]

[tex]= \frac{2.5\times 10^5(\frac{\lambda}{D_{eye}})}\frac{2.5\times 10^5(\frac{\lambda}{D_{telescope}})}[/tex]

[tex]\frac{\theta_{eye}}{\theta_{telescope}} = \frac{10}{8\times 10^{-3}} = 1250[/tex]

c) [tex]\theta_{eye} = 2.5\times 10^{5} \frac{5\times 10^{-7}}{7\times 10^{-2}}[/tex][tex]\theta_{eye} = 1.78\ arc\ second[/tex]

Final answer:

The diffraction limit of the human eye with a pupil diameter of 0.8 cm for 500 nm light is 7.6 × 10⁻⁵ radians. Compared to a 10-meter telescope, which has a diffraction limit of 6.1 × 10⁻⁸ radians, the telescope's resolution is significantly finer. If the human's two eyes acted as an optical interferometer, the limit would be approximately 8.7 × 10⁻⁶ radians.

Explanation:

Calculating the Diffraction Limit of the Human Eye and Comparison with a Telescope

A. To calculate the diffraction limit of the human eye, we can use the formula θ = 1.22 λ / D, where θ is the angle of resolution, λ is the wavelength of light, and D is the diameter of the lens or pupil. For the human eye with a wide-open pupil diameter of 0.8 cm and a light wavelength of 500 nm, the diffraction limit (θ) is approximately:

θ = 1.22 × 500 × 10⁻⁹ m / 0.008 m = 7.62 × 10⁻⁵ radians.

To express this answer with two significant figures, the diffraction limit is 7.6 × 10⁻⁵ radians.

B. For a 10-meter telescope with the same wavelength of light, the diffraction limit (θ) is calculated as:

θ = 1.22 × 500 × 10⁻⁹ m / 10 m = 6.1 × 10⁻⁸ radians,

which to two significant figures is 6.1 × 10⁻⁸ radians. This shows the much finer angular resolution of the telescope compared to the human eye.

C. Estimating the diffraction limit for human vision when considered as an 'optical interferometer' with two eyes 7 centimeters apart acting as one, we find that the effective diameter is now 7 cm instead of 0.008 m, and thus the diffraction limit (θ) is:

θ = 1.22 × 500 × 10⁻⁹ m / 0.07 m = 8.74 × 10⁻⁶ radians,

which to two significant figures is 8.7 × 10⁻⁶ radians.

What must be the pressure difference between the two ends of a 2.0-km section of pipe, 25 cm in diameter, if it is to transport oil?

Answers

flow rate F = .00087 m^3/s 
viscosity eta = 0.2 Pa-s 
pipe radius r = 0.105 m 
pipe length L = 2800 m 
dP = 8LF*eta/(pi*r^4) = 10206.8139295042 Pa

the pressure difference between the two ends of a 2.0-km section of pipe, 25 cm in diameter, if it is to transport oil 10206.8139295042 Pa

what is  viscosity ?

the viscosity can be defined as the resistance that a fluid will flow when sliding one sheet over another.

Kinematic viscosity can be defined as its unit only depend on kinematic units (m / s ^ 2) and not physical properties such as mass.

Viscosity is described as both liquids and gases, it refers to the ability of a gas or liquid to resist flow.

In other words, it  exists between the molecules of a fluid, which resists its flow.

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You are trying to push your stalled car. although you apply a horizontal force of 400 n to the car, it doesn't budge, and neither do you.

Answers

Three forces are stopping the car from moving, in this case, and they are all giving a force of 400N, as well. The road is exerting a frictional force of 400N on the person pushing the car, the car itself is also exerting a force of 400N, and the car is exerting a frictional force on the road, all at a measure of 400N.

Final answer:

The stalled car doesn't move because the applied force is balanced by static friction, resulting in no net force. Once moving, a constant speed implies balanced forces, and deceleration shows that opposing forces overcome the applied force. While push-starting, the resultant force is the sum of individual forces applied by each person.

Explanation:

When you apply a force to push your stalled car, and it doesn't move, it indicates that the resultant force acting on the car is zero. The force you are applying is balanced by the force of static friction between the car's tires and the ground, which prevents the car from budging. Additionally, there might be other factors such as the car's weight, potential mechanical resistance, or an incline which contribute to the car not moving. Applying a force of 400 N and not moving suggests that the static frictional force is equal to or greater than the applied force, resulting in no net force and, hence, no acceleration (according to Newton's Second Law).

For a car to move, the force applied must overcome the static friction. Once the car starts moving, if it continues at a constant speed, it suggests that the applied force is now balanced by kinetic friction and air resistance (drag force). If the car decelerates, this implies that the opposing forces (friction and drag) are greater than the applied force. To maintain a car's constant speed, the friction force resulting from the tires pushing against the road via the throttle, engine, and drive train should balance the air resistance.

If you and a friend attempt to push-start a stalled car, the combined horizontal forces you both apply (50 N and 45 N respectively) equal 95 N. This is found by simply adding the two forces, assuming they are applied in the same direction.

A car is driving northwest at v mph across a sloping plain whose height, in feet above sea level, at a point n miles north and e miles east of a city is given by h(n,e)=1500+75n+50e. (a) at what rate is the height above sea level changing with respect to distance in the direction the car is driving?

Answers

Refer to the diagram shown.

Given:
[tex]h(n,e) = 1500 + 75n + 50e[/tex]

Define 
[tex]\hat{r} = unit \, vector \, along \, \vec{v} \\ \hat{i} = unit \, vector \, east \\ \hat{j} = unit \, vector \, north \\ \nabla \equiv \hat{i} \frac{\partial}{\partial e} + \hat{j} \frac{\partial}{\partial n} [/tex]

[tex]\hat{r} = \frac{1}{ \sqrt{2} } (-\hat{i}+\hat{j} )[/tex]

Then the rate of change of h with respect to the vector v is
[tex]\nabla h . \hat{r} = \frac{1}{\sqrt{2}}(50\hat{i} + 75\hat{j}).(-\hat{i}+\hat{j}) = \frac{1}{\sqrt{2}} (-50+75) =17.68[/tex]

Answer: 17.7 ft per mile

What is the velocity of an object that has a mass of 2.5 kg and a momentum of 1,000 kg · m/s?

Answers

Using the momentum formula p=mv, it can be determined that v=p/m. Therefore 1000/2.5 = 400. The object is going 400m/s
The object's speed is (1000 kg-m/s) / (2.5 kg) = 400 m/s.

We can't describe its velocity because we don't know what direction it's moving.

A force of 160. N parallel to an inclined plane is required to move a 200. N weight up the inclined plane with a constant velocity. Find the coefficient of sliding friction if the plane is inclined at 30.0 degrees.

Answers

- Drawing a free body diagram will help you visualize the different forces acting on an object.

- Find ∑Fy and ∑Fx and solve for unknowns; force due to kinetic friction (fk) and normal force (Fn).

- After solving for unknowns, plug values into equation for kinetic friction, fk = μk • Fn, and solve for μk

Coefficient of kinetic/sliding friction = .3464

See attached photo for worked out solution.
Final answer:

The coefficient of sliding friction (µk) on the inclined plane described in the problem is approximately 0.923. This is calculated by equalizing the sliding friction force to the force necessary to move the object up the inclined plane, taking into account that their net force is zero because the object moves at constant velocity.

Explanation:

The subject of your question is related to physics, particularly to the section of mechanics that deals with friction and inclined planes. The inclined plane here introduces two dimensions of complexity since forces act both parallel and perpendicular to the plane.

Since the object is moving at constant velocity, the net force acting on it is zero. Therefore, the force necessary to move the object up the inclined plane, 160 N, is equivalent to the force of sliding friction.

We have to consider three forces acting on the object: the weight of the object (W = 200 N), the normal force (N), and the sliding friction force (f).

In this case, the normal force does not equal the weight of the object, as the inclined plane reduces the effective weight the object has in the perpendicular direction, as shown with these components:

N = W cos θ = (200 N) cos(30°) = 173.2 N

The sliding friction force (f) can be calculated using the formula f = µk N. As we established earlier, the force necessary to move the object up the inclined plane (160 N) is equivalent to the force of sliding friction. Hence:

160 N = µk x 173.2 N

µk = 160 N / 173.2 N

µk = 0.923

The answer to the problem is: The coefficient of sliding friction (µk) is approximately 0.923.

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A european car manufacturer reports that the fuel efficiency of the new microcar is 28.5 km/l highway and 22.0 km/l city. what are the equivalent fuel efficiency rates in miles per gal?]

Answers

We know that:

1 mile = 1.61 km

1 gal = 3.8 L

 

Therefore converting the fuel efficiency rates:

highway = (28.5 km/L) * (1 mile / 1.61 km) * (3.8 L / 1 gal) = 67.27 mile / gal

city = (22.0 km/L) * (1 mile / 1.61 km) * (3.8 L / 1 gal) = 51.93 mile / gal


An object is thrown directly up (positive direction) with a velocity (vo) of 20.0 m/s and do= 0. Determine how long it takes to get to the maximum height of 24.0 m.

Answers

t=d/v
t=24m/20m/s
t=6/5 s
t=1.2s

Answer:

It takes 2.04s to get to the maximum height

Explanation:

This is a vertical throw problem so it can be  treated as a uniform accelerated rectilinear motion. For computing time we are going to use the formula:

[tex]v_{f}=v_{o}+g*t[/tex]

where[tex]v_{f}[/tex] is the final velocity, [tex]v_{o}[/tex] is the initial velocity, [tex]t[/tex] is the time and, [tex]g[/tex] is the gravity.

For solving this kind of problems we need at least three values. The values we have are:

[tex]v_{o} = 20\dfrac{m}{s}[/tex][tex]g = -9.8\dfrac{m}{s^{2}} [/tex] (negative because gravity's direction is oposite from the object's moving direction)[tex]v_{f}=0[/tex] (final velocity equals zero because at maximun height the object stops moving)

Now:

[tex]v_{f}=v_{o}+g*t[/tex]

[tex]v_{f}-v_{o}=g*t[/tex]

[tex]\dfrac{v_{f}-v_{o}}{g}=t[/tex]

[tex]\dfrac{0-20}{-9.8}=t[/tex]

[tex]t=2.04s[/tex]

why does carrying furniture up four flights of stairs require twice as much work as caring furniture up two flights of stairs

Answers

because 4 flights of stairs is double to 2 fights of stairs therefore it is double the work

Final answer:

Carrying furniture up four flights of stairs requires twice as much work as carrying it up two flights due to the direct proportionality of work to the distance moved against gravity. The work done is calculated by multiplying the force by the distance over which the force is applied.

Explanation:

Work Done in Carrying Furniture Upstairs

Carrying furniture up four flights of stairs requires twice as much work as carrying it up two flights of stairs because work in physics is defined as the product of the force applied to an object and the distance over which that force is applied. In simpler terms, work is directly proportional to the distance. When you carry the furniture up four flights of stairs, you are moving it over twice the distance you would if you were carrying it up only two flights of stairs.

For example, if a traveler carries a 150 N suitcase up four flights of stairs for a total height of 12 m, the work done is equal to the force times the vertical distance (Work = Force × Distance). This can be calculated as Work = 150 N × 12 m = 1800 J. If the same suitcase were carried up only two flights of stairs with a total height of 6 m, the work done would be Work = 150 N × 6 m = 900 J, which is exactly half the work required to carry it up four flights.

Therefore, the amount of work doubles as the distance doubles, assuming the force remains constant.

Topographic maps represent an area's physical features by describing _____.
A. shape, elevation, and steepness of features
B. rock type, elevation, and shape of features
C. elevation and rock types of features and direction of river flow
D. shape, elevation, faults, and folds

Answers

a topographic map is a type of map characterized by large-scale detail and quantitative representation of reliefi would say a 

HELP PLEASE? :(

Which pair of atoms will form an ionic compound?
A) One atom of oxygen and two atoms of fluorine
B) One atom of calcium and two atoms of chlorine
C) One atom of nitrogen and three atoms of fluorine
D) Two atoms of nitrogen and four atoms of oxygen

Answers

Choice A, C, and D forms covalent bonds. Choice B is the only one hat has an ionic bond, making it the answer. I hope this helps.

The pair of atoms that will form an ionic compound is "One atom of calcium and two atoms of chlorine.". The correct option is B.

What is an ionic compound?

In an ionic compound, one atom gives up one or more electrons to another atom, resulting in positively charged cations and negatively charged anions that are attracted to each other due to electrostatic forces.

Here in the question

In option B, calcium (Ca) has two valence electrons, while chlorine (Cl) has seven valence electrons. To obtain a stable octet configuration, calcium will lose two electrons to form a Ca2+ cation, and two chlorine atoms will each gain one electron to form Cl- anions. The resulting compound, CaCl2, is an ionic compound with a crystal lattice structure held together by electrostatic forces between the oppositely charged ions.

Option A involves two non-metals, and they typically form covalent compounds, not ionic compounds.

Option C is similar to option A and also involves two non-metals, which typically form covalent compounds.

Option D involves two non-metals, and although the atoms can bond covalently, the compound formed would be a polar molecule, not an ionic compound.

Therefore, The correct option is B i. e One atom of calcium and two atoms of chlorine which forms an ionic compound.

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The distance of mars to the sun is 1.5 that of earth. "how many earth years does it take for mars to orbit the sun?"

Answers

4 years to orbit the sun



Yet another great question from our universe. Mars takes 686.971 earth days to orbit the sun. .
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