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Maksim231197 [3]
2 years ago
7

Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

eed 2v, and they feel no air resistance. Which statement about these balls is correct?A) At their highest point, the acceleration of each ball is instantaneously equal tozero because they stop for an instant.B) The balls will reach the same height because they have the same mass and thesame acceleration.C) At its highest point, ball B will have twice as much gravitational potential energyas ball A because it started out moving twice as fast.D) Ball B will go twice as high as ball A because it had twice the initial speed.E) Ball B will go four times as high as ball A because it had four times the initialkinetic energy.
Physics
1 answer:
MatroZZZ [7]2 years ago
6 0

Answer:

E) True.   Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

          Em₀ = K

          Em₀ = ½ m v²

Final

         Emf = U = mg h

         Em₀ = emf

        ½ m v² = mgh

        h = v² / 2g

For ball A

         h_A = v² / 2g

For ball B

        h_B = (2v)² / 2g

        h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False  has 4 times the gravitational potential energy than ball A

D) False.  It goes 4 times higher

E) True.

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A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?
Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, F_w↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, F_{(helicopter)}↑ = The weight of the crate = 2,452.5 N

The net force, F_{(net)} = F_{(helicopter)}↑ - F_w↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0
2 years ago
you want to compare brands of paper towels to see which holds the most liquid. the independent variable in your experiment would
Kazeer [188]

Answer: the brand of paper towel

Explanation: the independent variable is the one you control in an experiment. the dependent variable would be the amount of water in the paper towel

5 0
2 years ago
Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th
katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0
2 years ago
Read 2 more answers
4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that
Basile [38]

The strength of the magnetic field is 4.8\cdot 10^{-5} T

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

\epsilon = \frac{N\Delta \Phi}{\Delta t} (1)

where

N = 505 is the number of turns in the coil

\Delta \Phi is the change in magnetic flux through the coil

\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s is the time interval

\epsilon = 0.166 V

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

\Delta \Phi = B A cos \theta

where

B is the strength of the magnetic field

A is the area of the coil

\theta=0^{\circ} is the angle between the normal to the coil and the field

The area of the coil can be written as

A=\pi r^2

where

r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m is its radius

Substituting everything into eq.(1) and solving for B, we find:

\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

8 0
2 years ago
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The diagram shows the electric field due to point charge Q. Which statements are correct? Check all that apply.
Bess [88]

Answer:

The <em>correct</em> statements  are:

  • <em>A. The electric field is nonuniform.</em>
  • <em>D. Charge Q is positive.</em>
  • <em>E. If charge A moves toward charge Q, it must be a negative charge</em>

Explanation:

The answer choices are:

  • A. The electric field is nonuniform.
  • B. The electric field is uniform.
  • C. Charge Q is negative.
  • D. Charge Q is positive.
  • E. If charge A moves toward charge Q, it must be a negative charge.
  • F. If charge A moves toward charge Q, it must be a positive charge.

<h2>Solution</h2>

The <em>electric field</em> is the electrostatic force per unit of charge,  

         \vec E=\dfrac{\vec F}{Q}

around around a charge, where another charge would experience the electrostatic force.

The electric field lines are shown in a diagram with arrows ditributed radially away from a positive charge and radially toward a negative charge.

Since the arrows are away from Q, Q is a positive charge: <em>statement D.</em>

Since the size of the arrows decreases as you move away  from Q the stregth of the field is not uniform: <em>statement A.</em>

Since the charge Q is positive, a negative charge would be attracted toward it: <em>statement E.</em>

4 0
2 years ago
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