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Kitty [74]
3 years ago
13

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

Physics
1 answer:
Kamila [148]3 years ago
5 0

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

{v}^{2}  -  {u}^{2}  = 2gh

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

{30}^{2}  -  {0}^{2}  = 2 \times 10 \times h

=  > 20h = 900

=  > h =  \frac{900}{20}  = 45

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A(n) 930 N crate is being pushed across a level floor by a force of 400 N at an angle of 20◦ above the horizontal. The coefficie
Nana76 [90]

Answer:

The magnitude of the acceleration of the box is 2.291\,\frac{m}{s^{2}}.

Explanation:

The free body diagram of the crate is included as attachment, whose equations of equilibrium are described below:

\Sigma F_{x} = P\cdot \cos 20^{\circ} - \mu_{k}\cdot N = \left(\frac{W}{g}\right)\cdot a

\Sigma F_{y} = P\cdot \sin 20^{\circ} + N - W = 0

From second equation of equilibrium we find an expression for the normal force and find the respective value:

N = W - P \cdot \sin 20^{\circ}

N = 930\,N - 400\cdot \sin 20^{\circ}\,N

N = 793.192\,N

Lastly, the acceleration experimented by the crate during pushing is cleared in the first equation of equilibrium and consequently calculated:

a = \frac{P\cdot \cos 20^{\circ}-\mu_{k}\cdot N}{\frac{W}{g} }

a = \frac{400\cdot \cos 20^{\circ}\,N-0.20\cdot (793.192\,N)}{\frac{930\,N}{9.807\,\frac{m}{s^{2}} } }

a = 2.291\,\frac{m}{s^{2}}

The magnitude of the acceleration of the box is 2.291\,\frac{m}{s^{2}}.

3 0
3 years ago
If the distance between two charged particles is doubled, the force between them changes by a factor of ___. (1 point)
MatroZZZ [7]

Answer:

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3 years ago
For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot
Margarita [4]

Answer:

(a) -202 m/s²

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Explanation:

Given data

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a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²

(b) We can find the distance traveled (d) using the following kinematic expression.

y = v₀ × t + 1/2 × a × t²

y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²

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3 0
3 years ago
How does approaching a temperature of absolute zero affect kinetic energy.?
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If the temperature is high the colliding particles will collide more. and if the temperature is low the colliding particles will collide less.

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Absolute zero is the point where where all molecules have no kinetic energy. It is a theoretical value (it has never been reached).

The Kelvin temperature scale is based on absolute zero being the lowest possible temperature that could theoretically be reached. That is why there is no such thing as a negative Kelvin temperature value.

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