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Lelechka [254]
4 years ago
12

Darwin is sitting on top of a 10 m tree and decided to drop a 3 kg baseball to the ground. What is the velocity when the PE turn

s into KE?
Physics
1 answer:
Nimfa-mama [501]4 years ago
8 0

The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground.  It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.

It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.

More and more PE turns into KE as the ball falls, all the way down.

When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.

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3 years ago
Which statement is false?
Mkey [24]

Answer:

Once you reach adulthood, you will not experience peer pressure.

Explanation:

We tend to believe that self-control comes from within, but many of our attitudes depend as much on friends and family as on ourselves. That's because at any stage of our life (even the adult stage) our friends and our family have the power to influence us, sometimes that influence is good and sometimes bad, it's up to us to reason about them.

The people around us have the power to make us fat, consume more alcohol, worry less about the environment and expose ourselves to the sun without proper protection, among many other things.

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3 0
3 years ago
Read 2 more answers
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
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