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Travka [436]
3 years ago
7

In which situation is work not being done?

Physics
1 answer:
almond37 [142]3 years ago
5 0

AS

work done =W = F.d = F d cosФ     (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body.  F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example:  A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done  dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0   ( as cos 90°=0)

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Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

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  v = √(1.0(9.8))

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How fast must a 1000 kg car be moving to have a kinetic energy of:
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Answer:

\mathrm{(a)}\: 2\: \mathrm{m/s}\\\mathrm{(b)}\: 20\: \mathrm{m/s}\\

Explanation:

The kinetic energy of an object is given by KE=\frac{1}{2}mv^2 where m is the mass of the object and v is the velocity of the object.

We can set up the following equations with the information given:

\mathrm{(a)}\: 2.0\cdot 10^3=\frac{1}{2}\cdot 1000\cdot v^2, \\v=\fbox{$2\: \mathrm{m/s}$}

For part B, we have the same equation, but kinetic energy is now 2.0\cdot 10^5.

Therefore:

\mathrm{(b)}\: 2.0\cdot 10^5=\frac{1}{2}\cdot 1000\cdot v^2, \\v=\fbox{$20\: \mathrm{m/s}$}.

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3 years ago
What charge does aluminum have?
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Estimate the speed of the water free surface and the time required to fill with water a cone-shaped container 1.5 m high and 1.5
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Answer:

speed of water is 0.0007138m/s

Explanation:

From the law of conservation of mass

Rate of mass accumulation inside vessel = mass flow in - mass flow out

so, dm/dt = mass flow in - mass flow out

taking p as density

d \frac{dQ}{dt} = pq_i_n

where,

q(in) is the volume flow rate coming in

Q = is the volume of liquid inside tank at any time

But,

dQ = Adh

where ,

A = area of liquid surface at time t

h = height from bottom at time t

A = πr²

r is the radius of liquid surface

r = (1.5/2) \div 1.5 h = \frac{h}{2}

Hence,

\pi( \frac{h}{2} )^2\frac{dh}{dt} =q_i_n

\frac{dh}{dt} = \frac{q_i_n}{\pi (\frac{h}{2})^2 } =\frac{4q_i_n}{\pi h^2}

so, the speed of water surface at height h

v = \frac{dh}{dt} =\frac{4q_i_n}{\pi h^2}

where,

q_i_n is 75.7 L/min = 0.0757m³/min

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v = \frac{4 \times 0.0757}{\pi \times 1.5^2} \\\\v = 0.04283m/min

v = 0.04283 /60

v = 0.0007138m/s

Hence, speed of water is 0.0007138m/s

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3 years ago
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