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FrozenT [24]
3 years ago
7

3. A block with a force of the Earth of -300. N is moved at constant velocity over a horizontal surface by a force of +50.0 N ap

plied parallel to the surface. What is the coefficient of kinetic friction?​
Physics
1 answer:
KIM [24]3 years ago
3 0

Answer:

-6

Explanation:

FRICTIONAL force=-300N,NORMAL REACTION,R=+50.0

Ú=F/R

=-300/+50.0

=-6

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Since the wire is not splitting at any point in the circuit,

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Equivalent Resistance = 60 Ω

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cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?
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4 years ago
The fastest time ever run at Pikes Peak is 7:57. That’s way too long. If we removed all the curves, making a straight line from
anygoal [31]

Answer: 271.4 s

Explanation:

We are told the top speed (maximum speed) V_{max} the car has is:

V_{max}=203 mph=90.74 m/s taking into account 1 mile=1609.34 m

And the car's average acceleration a_{ave} is:

a_{ave}=0.091 g=2.93 ft/s^{2}=0.89 m/s^{2}

Since:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t} (1)

Where:

V_{f}=V_{max}=90.74 m/s is the car's final speed (top speed)

V_{o}=0 m/s because it starts from rest

\Delta t is the time it takes to reach the top speed

Finding this time:

\Delta t=\frac{V_{f}-V_{o}}{a_{ave}} (2)

\Delta t=\frac{90.74 m/s - 0 m/s}{0.89 m/s^{2}} (3)

\Delta t=t_{1}=101.95 s (4)

Now we have to find the distance d the car traveled at this maximum speed with the following equation:

V_{f}^{2}=V_{o}^{2} + 2a_{ave} d (5)

Isolating d:

d=\frac{V_{f}^{2}}{2a_{ave}} (6)

d=\frac{(90.74 m/s)^{2}}{2(0.89 m/s^{2})} (7)

d=4625.70 m (8)

On the other hand, we know the total distance D traveled by the car is:

D=12.42 miles = 19988.052 m

Hence the remaining distance is:

d_{remain}=D-d=19988.052 m - 4625.70 m (9)

d_{remain}=15362.35 m (10)

So, we can calculate the time t_{2} it took to this car to travel this remaining distance d_{remain} at its top speed V_{max}, with the following equation:

V_{max}=\frac{d_{remain}}{t_{2}} (11)

Isolating t_{2}:

t_{2}=\frac{d_{remain}}{V_{max}} (12)

t_{2}=\frac{15362.35 m}{90.74 m/s} (13)

t_{2}=169.45 s (14)

With this time t_{2} and the value of t_{1} calculated in (4) we can finally calculate the total time t_{TOTAL}:

t_{TOTAL}=t_{1}+ t_{2} (15)

t_{TOTAL}=101.95 s + 169.45 s (16)

t_{TOTAL}=271.4 s s

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