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velikii [3]
3 years ago
5

1. a body of mass 40kg is given an acceleration of 10ms/2 on a horizontal ground for which the coefficient of friction 0.5, calc

ulate the force required to accelerate the body (g= 10ms/2)
2. a force of 20N applied parallel to the surface of a horizontal table and just sufficient to make a block of mass 4kg move on the table. Calculate the coefficient of friction between the block and the tabla.​
Physics
1 answer:
vodka [1.7K]3 years ago
5 0

Answer:

1: Force, F = 200N

2: coefficient of friction = 0.5

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A. Energy was used to change phase of water from liquid to gas (evaporation). it is radiated back into the atmosphere from the stove.
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I notice the wheels on my bicycle spin in a circular motion as I am pedaling. I know that the
kherson [118]

Answer:

Acceleration, a = 750m/s²

Explanation:

Given the following data;

Radius, r = 0.31m

Velocity, v = 15m/s

To find the centripetal acceleration;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Acceleration, a = \frac {15^{2}}{0.31}

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Acceleration, a = 750m/s²

Therefore, the centripetal acceleration of the bike wheel is 750m/s².

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3 years ago
Let's say you are on the third story of an apartment building that has a swimming pool. For some insane reason you think it is a
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Answer:

To make it into the pool you must run and jump at

\displaystyle v_o=x.\sqrt{\frac{g}{2y}}

Explanation:

Horizontal Launch

When an object is thrown with a specified initial speed in the horizontal direction, it describes a curved path that finishes when it hits the ground level after traveling certain horizontal distance x and a vertical height y from the launching point. The horizontal speed is always constant and the vertical speed increases due to the effect of gravity. It can be found that the horizontal distance reached by the object when launched at an initial speed  in a given time t is

x=v_o.t

And the vertical distance is

\displaystyle y=\frac{g.t^2}{2}

If t is the total flight time, then x and y are maximum and we can find a relation between them. Solving for t in the first equation

\displaystyle t=\frac{x}{v_o}

Substituting in the second equation

\displaystyle y=\frac{g}{2}\left ( \frac{x}{v_o}\right )^2

Rearranging

\displaystyle \left ( \frac{v_o}{x}\right )^2=\frac{g}{2y}

Solving for v_o

\displaystyle v_o=x.\sqrt{\frac{g}{2y}}

There are many applications for the horizontal launch. One common situation is when someone wants to drop something on certain terrain at a specific approximate point when traveling in a plane at a given height. Once the object is left fall, it has the same speed as the plane, so the plane speed can be estimated to make the best possible launch, or given that speed, we can know in advance where the object will reach ground level

4 0
3 years ago
What is the acceleration of a 7 kg mass pushed by a 3.5 N force?
defon

Answer:

0.5m/s²

Explanation:

F =m x a

3.5 = 7 x a

a = 3.5 /7

a = 0.5m/s ²

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What is the potential energy of a 1000 kg-ball that is on the ground?
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Answer:

0J

Explanation:

PE=mgh

PE=  1000kgx9.8m/s^2x0m

PE = 0J

6 0
2 years ago
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