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Dahasolnce [82]
3 years ago
5

Your cousin is moving into an apartment in San Francisco. This apartment is on a street that is angled at 25∘ above the horizont

al. He places a box of mass 15kg on the street. The coefficients of friction between the box and the street are μs=0.45 and μk=0.25.
A: He releases the box after he places it on the ground. Will it start to slide down the sloped road?
Yes or No?
B: As the box slides down the slope, what is its acceleration?
C: What is the magnitude of the minimum force he would have to apply to the box to keep it from sliding down the hill? Assume that he pushes parallel to the road and wants to hold it in place.
D: What is the magnitude of the minimum force he would have to apply to start pushing the box up the hill? Again assume that he pushes parallel to the road.
Physics
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

70

Explanation:

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just olya [345]

Answer:

3.75*10^{-13}  electrons

Explanation:

The total charge Q is the sum of the charge of the N electrons contained in the sphere:

Q=N*q_{e}

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We solve to find N:

N=\frac{Q}{q_{e}}=\frac{-6 x 10^{-6}}{-1.6 x 10^{-19}}=3.75*10^{-13}

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3 years ago
The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus
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Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

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P_{f} -P_{i}=F*delta(t)

2m(v_{f} -v_{i})=F*delta(t)

v_{i} =0.35-0.35=0

It is neccesary calculate the force:

F=(m+m)*g*sin\theta

Here, g = gravity = 9.8 m/s²

F=(0.2+0.2)*9.8*sin37=2.3591N

v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s

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