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mr_godi [17]
3 years ago
7

A person jumps off a diving board 5.0 m above the water's surface into a deep pool. The person's downward motion stops 2.2 m bel

ow the surface of the water. Calculate the speed of the diver just before striking the water, in m/s (ignore air friction).
Physics
1 answer:
shutvik [7]3 years ago
7 0

Answer:

9.9 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s

If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.

Hence, the speed of the diver just before striking the water is 9.9 m/s

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A boy starts from point A and walks 3 meters toward the north, then turns around and walks 6 meters toward the south.
docker41 [41]

Answer:

Total distance traveled = 9 m

Explanation:

Given:

Distance travel towards north = 3 meter

Distance travel towards south = 6 meter

Find:

Total distance traveled

Computation:

Total distance traveled = Sum of total distance

Total distance traveled = Distance travel towards north + Distance travel towards south

Total distance traveled = 3 m + 6 m

Total distance traveled = 9 m

3 0
2 years ago
A force acts on a body of mass 13 kg initially at rest. The force
PtichkaEL [24]

Answer:

Force that acted on the body was F = 13 N

Explanation:

If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s

When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

v_f=v_i+a*t

which in our case becomes;

10\,m/s=0+a*(10\,s)

and we can solve for the acceleration as:

a = 10/10  m/s^2 = 1 m/s^2

Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:

F = 13 kg * 1 m/s^2 = 13 N

4 0
3 years ago
A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat
KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

L_s = Latent heat of steam

s_w = Specific heat of water

L_i = Latent heat of ice

v = Velocity of ice

\Delta T = Change in temperature

Amount of heat required for steam

Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)

Heat released from water at 100 °C

Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6

Heat released from water at 0 °C

Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)

Total heat released is

Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m

The kinetic energy of the bullet will balance the heat

K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s

The velocity of the ice would be 2452.79432 m/s

6 0
3 years ago
A rocket is fired at 100 m/s at an angle of 37, what was its speed at the top of its path?
Nadusha1986 [10]

Answer:

0m/s

Explanation:

Since its fired at an angle, at the top there will be a split second where the velocity will be 0, as it has a parabolic shape, so the speed at the top of its path is 0

4 0
2 years ago
An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its
Sidana [21]
Internal energy, U, is equal to the work done or by the system, plus the heat of the system: <span>ΔU=q+w
</span>in the question they tell you the work done by the system, and the internal energy:
8185 J= -346 J + q work is negative because it was done BY the system.
substitute in: <span>q=m∗Cp∗ΔT</span> and solve for <span>Cp</span><span>.
</span>
-------------------------------------
remember that <span>ΔT=<span>Tf</span>−<span>Ti
</span></span>
so the equation, really, is: <span>q=m∗Cp∗(<span>Tf</span>−<span>Ti</span>)</span><span>
------------------------------------------
</span>
<span>185J=−346J+[m∗Cp∗(<span>Tf</span>−<span>Ti</span>)]

</span>plug in the rest of your values and solve for <span><span>Cp</span></span>
4 0
3 years ago
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