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Ray Of Light [21]

3 years ago

11

There are many forms of energy. All energy falls into one of two categories, either potential or kinetic energy. A Newton's Crad

le, with the pendulums at different points along its path, involves kinetic energy being transformed into potential energy. Use scientific reasoning to explain in detail how the kinetic is transformed into potential energy in the Newton's Cradle example. Be sure to explain the differences and similarities between kinetic and potential energy in your response.

1 answer:

-Dominant- [34]3 years ago

3 0

Answer:

no motion

Explanation:

kinetic energy transformed to potential energy when no movement or motion..potential is more on gravity

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After the same car leaves the platform, gravity causes it to accelerate downward at a rate of 9.8 m/s2. What is the gravitationa

Simora [160]

The gravitational force on the car is

(9.8 m/s^2) x (the car's mass in kg).

The unit is newtons.

7 0

3 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

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3 years ago

If Shirley pushes one end of a desk to the right with a force of 12.0N and anfernee pushes the desk at the same time to the left

Marianna [84]

The net force does not depend on the mass.

We have 12N to the right, and 19N to the left.
The net force is (19.0-12.0)N=7.0N to the left.

5 0

4 years ago

If you drop a rock off a cliff and it hits the ground 3 seconds later how tall is that cliff

skelet666 [1.2K]

Answer:

Basic kinematics, negating drag and assuming ideal conditions, we use the equation:

d=vi*t+1/2*a*t^2

Since vi is 0 (we know this because you’re dropping it, not throwing it)…

…and the only acceleration acting on it is gravity, a=9.8 m/s^2…

…we get

d=1/2(9.8)(5)^2

Explanation:

Some quick mental math tells us that this is about 125 m.

Plugging it in, we find it to be 122.5 m.

6 0

3 years ago

The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb

topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

$\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\$

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

$I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2$

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

$F = \frac{mv^2}{r}$

The relation between linear velocity and the angular velocity is

$v = \omega r$

So,

$F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}$

As can be seen, the maximum velocity for the projectile without breaking the cable is $\sqrt{1500r}$ , where r is the length of the cable.

6 0

3 years ago