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DedPeter [7]
2 years ago
7

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

of static friction f on the block?
Physics
2 answers:
galina1969 [7]2 years ago
6 0

Answer:

μ =tanθ

Explanation:=

The ratio of the force of static friction and the normal reaction is equal to tanθ. F=mgsinθ. R = mgcosθ.

μ=tanθ

Aliun [14]2 years ago
5 0

Answer: So the static friction force must be equal to g*m*cos( θ)

Explanation: The force downwards of the block will be equal to the component with respect of where it can move.

So the force downwards will be:

F = Weight*cos( θ)

And we know that the weight is w = -m*g where g is the gravitational acceleration.

F = -g*m*cos( θ)

And we know that the block is at rest, so the net forces on it are equal to zero, this means that the gravitational force must be canceledby other one, and this can be the static friction force.

So the static friction force must be equal to g*m*cos( θ)

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A 0.200 H inductor is connected in series with a 88.0 Ω resistor and an ac source. The voltage across the inductor is vL=−(12.0V
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Answer:

a.  (VL)R/ωL[1 - cos[ωt]]  = (10.84 V)[1 - cos[(487rad/s)t]]

b. 1.084 mV

Explanation:

a. Since it is a series circuit, the current in the inductor is the same as the current in the resistor.

Now, the voltage across the inductor vL = -Ldi/dt.

So, the current, i = -1/L∫vLdt.

Now, vL = −(12.0V)sin[(487rad/s)t] and L = 0.200 H

Substituting these into i, we have

i = -1/L∫vLdt

= -1/0.200H∫[−(12.0V)sin[(487rad/s)t]]dt.

= -[−(12.0V)]/0.200H∫[sin[(487rad/s)t]]dt.

= 60V/H∫[sin[(487rad/s)t]]dt

Integrating i, we have

i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + C

at t = 0, i(0) = 0

0 = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)× 0]] + C

0 = 60V/H ÷ [(487rad/s)[-cos[0]] + C

0 = 60V/H ÷ [(487rad/s)[-1]+ C

C = 60V/H ÷ [(487rad/s)

So, i = 60V/H ÷ [(487rad/s)[-cos[(487rad/s)t]] + 60V/H ÷ [(487rad/s)

i =  60V/H ÷ [(487rad/s)[1 - cos[(487rad/s)t]]

i = (0.123A)[1 - cos[(487rad/s)t]] = VL/ωL[1 - cos[ωt]] where ω = 487rad/s and VL = 12.0 V and L = 0.200 H

So, the voltage across the resistor vR = iR where R = resistance of resistor = 88.0 Ω

So, vR = iR = VL/ωL[1 - cos[ωt]] × R = (VL)R/ωL[1 - cos[ωt]]

=  (0.123A)[1 - cos[(487rad/s)t]] × 88.0 Ω

= (10.84 V)[1 - cos[(487rad/s)t]]

b. vR at t = 2.00 ms = 0.002 s

So, vR = (10.84 V)[1 - cos[(487rad/s)(0.002)]]

= (10.84 V)[1 - cos[0.974]]

= (10.84 V)[1 - 0.9999]

= (10.84 V)(0.0001)

= 0.001084

= 1.084 mV

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