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kifflom [539]
2 years ago
13

A ball is tossed in the air and released. It moves up, reverses direction, falls back down again, and is caught at the same heig

ht it was released. Considering the time interval after the ball is released and before it is caught, when does the gravitational potential energy of the ball have its maximum value?
Physics
2 answers:
PolarNik [594]2 years ago
8 0

Answer:

The potential energy has a  maximum when the ball is a time that is half of the time for total travel

Explanation:

Generally potential energy is a the varies directly with the height according to this formula

            PE =mgh

and the ball attains a maximum height when the time is equal to half of the total time taken to travel  

Anni [7]2 years ago
3 0

Answer:

The potential energy is highest at it point of Max height...

Explanation:

From the formula potential energy = mgh...

The greater the height, the greater the potential energy.

So considering the time interval it's at the time it takes the ball to reach the reversal point ( its Max height)

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A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

t=4.05 s

4 0
2 years ago
Suppose the truck that’s transporting the box In Example 6.10 (p. 150) is driving at a constant speed and then brakes and slows
Scorpion4ik [409]

Answer:

Friction acts in the opposite direction to the motion of the truck and box.

Explanation:

Let's first review the problem.

A moving truck applies the brakes, and a box on it does not slip.

Now when the truck is applying brakes, only it itself is being slowed down. Since the box is slowing down with the truck, we can conclude that it is friction that slows it down.

The box in the question tries to maintains its velocity forward when the brakes are applied. We can think of this as the box exerting a positive force relative to the truck when the brakes are applied. When we imagine this, we can also figure out where the static friction will act to stop this positive force. Friction will act in the negative direction. Or in other words, friction will act in the opposite direction to the motion of the truck and box. This explains why the box slows down with the truck, as friction acts to stop its motion.

5 0
2 years ago
Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o
ipn [44]

Answer:

U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

                              E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

             ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

                              E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

                              E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

                              E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence,                   E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

                               U = E_p + E_k + E_g

                               U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0
2 years ago
Consider three drinking glasses. All three have the same area base, and all three are filled to the same depth with water. Glass
Kay [80]

The glass which has the greatest liquid pressure at the bottom is all 3 have equal non-zero pressure at the bottom. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query about and it will be able to help you.

4 0
2 years ago
Read 2 more answers
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
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