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Verizon [17]
3 years ago
6

During a certain time interval, a constant force delivers an average power of 4 watts to an object. if the object has an average

speed of 2 meters per second and the force acts in the direction of motion of the object, what is the magnitude of the force f~ ?
Physics
2 answers:
bija089 [108]3 years ago
3 0
<span>Force = Work done / distance = 4Nm / 2m = 2N</span>
ivolga24 [154]3 years ago
3 0

Answer:

2 N

Explanation:

The average power is defined as:

P=\frac{W}{\Delta t}(1)

Here W is the work done and t is the time interval in which the work is done.

The average speed is the rate of change of the position with respect to time:

v=\frac{x}{\Delta t}\\\Delta t=\frac{x}{v}(2)

Replacing (2) in (1):

P=\frac{Wv}{x}\\x=\frac{Wv}{P}(3)

The work done by a force F acting on an object undergoing a distance x, is given by:

W=Fxcos\theta

the force acts in the direction of motion of the object, so the angle between then is zero and the cosine is equal to one:

x=\frac{W}{F}(4)

Equaling (3) and (4):

\frac{Wv}{P}=\frac{W}{F}\\F=\frac{P}{v}\\F=\frac{4W}{2\frac{m}{s}}\\F=2N

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A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
galina1969 [7]

Answer:

\mu_s = 0.62

\mu_k = 0.415

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

F = ma\\F = 1.7m

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

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3 years ago
A solar-powered car is traveling at constant speed around a circular track. what happens to the centripetal acceleration of the
anyanavicka [17]
Centripetal acceleration is the acceleration of an object acted upon a force that pulls it toward the center of origin. For constant speed, there is no acceleration, since acceleration is defined as the difference of velocities at an elapsed time. Hence, if speed doubles, then the centripetal acceleration would increase as well.
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How does the sun's heat affect the humidity of a place​
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Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli
Masja [62]

Answer:

Not possible

Explanation:

E_{cl} = longitudinal modulus of elasticity = 35 Gpa

E_{ct} = transverse modulus of elasticity = 5.17 Gpa

E_m = Epoxy modulus of elasticity = 3.4 Gpa

V_{\rho l} = Volume fraction of fibre (longitudinal)

V_{\rho t} = Volume fraction of fibre (transvers)

E_f = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764

Transverse modulus of elasticity is given by

E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148

V_{\rho l}\neq V_{\rho t}

Hence, it is not possible to produce a continuous and oriented aramid fiber.

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3 years ago
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Its called compression
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