What is the efficiency of the machine if the input is 263 J and the output is 142 J?

A student drops a rock from a bridge to the water 12m below. How many seconds does it take the rock to hit the water?

ANEK [815]

R = $\frac{g t^{2} }{2}$
r - displacement or height
t - time taken
g = 10 m/s-2
t = square root from 2r/g = $\sqrt{ \frac{2 x 12}{10} }$ = 1.5 seconds

7 0

3 years ago

What happens to the heat energy when you increase the length of an object

Talja [164]

Answer: When the temperature of an object increases, the average kinetic energy of its particles increases. When the average kinetic energy of its particles increases, the object's thermal energy increases. Therefore, the thermal energy of an object increases as its temperature increases.

8 0

3 years ago

While fishing on a lake, a fisherman notices ripples 0.32 m apart, and they are

Karo-lina-s [1.5K]

the main properties of the main wave propertioes include wavelength amplitude, cruest an trough

5 0

2 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

3 0

3 years ago

Where could convection currents form? Check all that apply.

KatRina [158]

<h3>Answer;</h3>

In a freshwater lake
In the atmosphere
In Earth's mantle

<h3>Explanation;</h3>

Convection currents are types that cause the process of convection, which the transfer of heat energy that occurs in fluids.
Convection currents are circular patterns that occurs in fluids such that the less dense warm fluids rises up while denser cold fluids sinks, it is this movement of less dense warm fluid and denser cold fluids that creates circular patterns that causes the process of convection to take place.
Convection currents may occur in the atmosphere where warm air rises while cold denser air sinks or moves towards the bottom, it may also occur in the mantle of the Earth and water or water bodies such as lakes.

8 0

3 years ago

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