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mote1985 [20]
2 years ago
14

A 50-kg block is at rest on a 15o slope. A force of 250 N is acting on the block up the slope parallel to it. If the block does

not slide up the slope, what is the minimum value of the coefficient of static friction between the block and the slope
Physics
1 answer:
Iteru [2.4K]2 years ago
6 0

The minimum value of the coefficient of static friction between the block and the slope is 0.53.

<h3>Minimum coefficient of static friction</h3>

Apply Newton's second law of motion;

F - μFs = 0

μFs = F

where;

  • μ is coefficient of static friction
  • Fs is frictional force
  • F is applied force

μ = F/Fs

μ = F/(mgcosθ)

μ = (250)/(50 x 9.8 x cos15)

μ = 0.53

Thus, the minimum value of the coefficient of static friction between the block and the slope is 0.53.

Learn more about coefficient of friction here: brainly.com/question/20241845

#SPJ1

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Explanation:

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2 years ago
Two football players are pushing a 60 kg blocking sled across the field at a constant speed of 2.5 m/s. The coefficient of kinet
expeople1 [14]

Answer:

1.06 m

Explanation:

The kinetic friction of the grass on the sled is the product of friction coefficient and the normal force, which equals to gravitational force. Let g = 9.81 m/s2:

F_f = \mu N = \mu mg = 0.3*60*9.81 = 176.58 N

So the deceleration caused by kinetic friction on the 60 kg sled is

a = F_f / m = 176.68 / 60 = 2.943 m/s^2

If the sled on a speed of 2.5m/s and then subjected to a deceleration of 2.943 m/s2, then we can use the following equation of motion to find out the distance traveled by the sled before rest:

v^2 - v_0^2 = 2a\Delta s

where v = 0 m/s is the final velocity of the sled when it stops, v_0 = 2.5m/s is the initial velocity of the can when it hits, a = -2.943 m/s2 is the deceleration, and \Delta s is the distance traveled, which we care looking for:

0^2 - 2.5^2 = 2*2.943\Delta s

\Delta s = 2.5^2 / (2*2.943) = 1.06 m

4 0
3 years ago
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A vertical wire carries a current vertically upward in a region where the magnetic field vector points toward the north. What is
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Answer:

West

Explanation:

We can solve the problem by using the right-hand rule. We can apply the rule as follows:

- Index finger: direction of the current

- middle finger: direction of the magnetic field

- thumb: direction of the force exerted on the wire

By applying the rule to this situation, we have:

- index finger: upward (current)

- middle finger: north (magnetic field)

- thumb: west (force)

So, the direction of the force exerted on the wire is to the west.

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What is a dog with four legs
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A dog with four legs is an intact, marginally redundant, canine quadruped.
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The circuit in Fig. P4.23 utilizes three identical diodes having IS = 10−14 A. Find the value of the current I required to obtai
Sliva [168]

Answer:

v = \frac{2 V}{3}= 0.667 v

Since we have identical diodes we can use the equation:

I_D =I= I_S e^{\frac{V_D}{V_T}}

And replacing we have:I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA

Since we know that 1 mA is drawn away from the output then the real value for I would be

I_D = I = 3.86 mA -1 mA= 2.86 mA

And for this case the value for v_D would be:

V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V

And the output votage on this case would be:

V = 3 V_D = 3 *0.660 V = 1.98 V

And the net change in the output voltage would be:

\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode v_1 + v_2 + v_3= 2 and each voltage is the same v for each diode, so then:

v = \frac{2 V}{3}= 0.667 v

Since we have identical diodes we can use the equation:

I_D =I= I_S e^{\frac{V_D}{V_T}}

And replacing we have:

I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA

Since we know that 1 mA is drawn away from the output then the real value for I would be

I_D = I = 3.86 mA -1 mA= 2.86 mA

And for this case the value for v_D would be:

V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V

And the output votage on this case would be:

V = 3 V_D = 3 *0.660 V = 1.98 V

And the net change in the output voltage would be:

\Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V

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2 years ago
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