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Strike441 [17]
2 years ago
7

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

k in half as much time?
A) 12P
B) 6P
C) P
D) P
E) P
Physics
1 answer:
schepotkina [342]2 years ago
4 0

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

P=\frac{W}{T}

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

P=\frac{W}{T}

Later, the force does six times more work, so the work done now is

W'=6W

And this work is done in half the time, so the new time is

T'=\frac{T}{2}

Substituting into the equation of the power, we find the new power produced:

P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P

So, 12 times more power.

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Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is 6.25\times 10^{- 3} ft/s^{2}

Solution:

As per the question:

Constant velocity of the horse in the horizontal, v_{x} = 1 ft/s

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

20^{2} + 10^{2} = l^{2}

l^{2}= 500

Now,

x^{2} + y^{2} = 500                            (1)

Differentiating equation (1) w.r.t 't':

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0

x\frac{dx}{dt} = - y\frac{dy}{dt}

where

\frac{dx}{dt} = Rate of change of displacement along the horizontal

\frac{dy}{dt} = Rate of change of displacement along the vertical

v_{x} = velocity along the x-axis.

v_{y} = velocity along the y-axis

xv_{x} = -yv_{y}

v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s

|v_{y}| = 0.5\ ft/s

Acceleration of the hay bale is given by the kinematic equation:

v_{y}^{2} = u_{y} + 2ay

(-0.5)^{2} =0 + 2ay

0.25 = 2ay

\frac{0.25}{2y} = a

a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}

7 0
3 years ago
A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
Flura [38]

Answer:

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Explanation:

Given:

Radius r = 0.330

m

Force F = 290 N

Angular acceleration \alpha  = 0.814 \frac{rad}{s^{2} }

From the formula of torque,

 Γ = I\alpha                                        (1)

 Γ = rF                                       (2)

rF = I \alpha

Find momentum of inertia I from above equation,

I = \frac{rF}{\alpha }

I = \frac{0.330 \times 290}{0.814}

I = 117.56 Kg. m^{2}

Find the momentum inertia of disk,

 I = \frac{1}{2}  Mr^{2}

M = \frac{2I}{r^{2} }

M = \frac{2 \times 117.56}{(0.330)^{2} }

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olganol [36]
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Answer:

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Explanation:

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