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3 years ago

Gravitational potential energy = the energy due to an objects

Physics

1 answer:

Vadim26 [7]3 years ago

5 0

Answer:

Gravitational energy is the energy of a body when it is taken to a certain height with respect to the earth surface.It is one of the part of potential energy that comes in thr part of the mechanical energy. the work done in taking a body upto a certain height get stored in the body in form of gravtitational potential energy. potential energy is directly proportional to height or altitude upto which the body is taken, mass of the body and gravtational acceleration(9.8m/s^2)So P.E. = mgh

Hope it helps

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Problem 28.3 Suppose that you are planning a trip in which a spacecraZ is to travel at a constant velocity for exactly six month

iragen [17]

Answer:

v = 0.999981c m/s

Explanation:

Using the time dilation equation

$T = \frac{T_{0} }{\sqrt{1 - \frac{v^{2} }{c^{2} } } }$

T = stationary time = 100 years

T₀ = 11/12 years = 0.917 years

v = speed of travel in the space = ?

c = speed of light = 3 * 10⁸ m/s

$100 = \frac{0.917 }{\sqrt{1 - \frac{v^{2} }{(3*10^{8} )^{2} } } }\\$

$(0.917/100) ^{2} = 1 - \frac{v^{2} }{ 9 * 10^{16} }$

v = 299987395.57 m/s

v = 2.99 * 10⁸ m/s

v = 0.999981c m/s

3 0

3 years ago

When observing a group of children at a daycare center, Emily made the following observations: Five year old children played in

konstantin123 [22]

This question is not about physics science.

The answer is: option <span>a. Five-year-old children have longer attention spans than three-year-old children.

It is the attention ability what let the older children to stay longer in one location instead of being moving between different activities. The younger children who cannot keep their attention long time in a same activity entertain themselves by changing activities.
</span>

5 0

3 years ago

Read 2 more answers

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

intial velocity is 45km/h, final velocity is 0, time taken is 3 seconds,calculate the retradation of the vehicle and distance tr

V125BC [204]

Answer:

v= u+at

0=45×1000×60×60+3a

_3a=16200045

a=5400015m per sec

4 0

3 years ago

Read 2 more answers

A sound wave has a frequency of 741 hz in air a wavelength of 0.47m what is the temperature of the air as soon as the below city

nekit [7.7K]

Answer:

Temperature when sound wave with wavelength 0.47m has frequency 741 hz is 188 degrees Centigrade.

Step-by-Step Explanation:

Given:

Speed of sound at 0 degrees centigrade = 235 ms

frequency = f = 741 hz

wavelength = w = 0.47m

speed of sound wave = wavelength * frequency = 741 * 0.47 = 348.27 m/s

Using the formula for speed of sound at a specific temperature:

Speed of sound at T degrees Centigrade = Speed of sound at 0 degrees + 0.6 * T

When speed of sound is 348 m/s, the temperature is given by:

348 = 235 + 0.6(T)

T = (348-235)/0.6

T = 188 degrees Centigrade

6 0

3 years ago