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3 years ago

Explain the conversion of energy in a roller coaster. Explain how energy is converted

1 answer:

5 0

As the rollercoaster goes up. kinetic energy changes to gravitational potential energy. When it moves back down, gpe changes back to ke.

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A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne

gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

$\Delta p = m \Delta v= m (v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

$v=-u$

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

$\Delta p=m(v-u)=m((-u)-u)=-2mu$

- For the clay, the final velocity is

$v=0$

since it sticks to the wall. So, the change in momentum is

$\Delta p = m(v-u)=m(0-u)=-mu$

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0

3 years ago

Rita is a salon owner. She notices that her salon charged one of her clients, Linda, extra for a service that the clent did not

Aloiza [94]

Answer:

C hope it helps

call the client and inform her that she was incorrectly charged.

7 0

3 years ago

What is the approximate efficiency of the engine?

Elodia [21]

Answer:91% I think

Explanation:

8 0

3 years ago

Read 2 more answers

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?

fomenos

Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

Explanation:

7 0

2 years ago