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poizon [28]
3 years ago
14

What is the maximum speed at which a car can safely travel around a circular track of radius 75.0 m if the coefficient of fricti

on between the tire and road is 0.200?
A) 3.87 m/s
B) 12.1 m/s
C) 15.0 m/s
D) 147 m/s
Physics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

B) 12.1 m/s

Explanation:

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the radial direction:

∑F = ma

F = m v² / r

Nμ = m v² / r

Substituting and solving for v:

mgμ = m v² / r

gμ = v² / r

v = √(gμr)

Given that μ = 0.200 and r = 75.0 m:

v = √(9.81 m/s² × 0.200 × 75.0 m)

v = 12.1 m/s

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Answer:

The specific gravity of the unkown liquid is 15.

Explanation:

Gauge pressure, at the bottom of the tank in this case, can be calculated from

P_{bot}=\gamma_{oil}h_1 + \gamma_{unk}h_2,

where h_1 and h_2 are the height of the column of oil and the unkown liquid, respectively. Writing for \gamma_{unk}, we have

\gamma_{unk} = \frac{P_{bot} - \gamma_{oil}h_1}{h_2} = \frac{65\frac{kN}{m^2} - 8.5\frac{kN}{m^3}\times 5m}{1.5m} = \mathbf{15 \frac{kN}{m^3}}.

Relative to water, the unknow liquid specific weight is 15 times bigger, therefore this is its specific gravity as well.

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2 years ago
The mass of the car?
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Answer:

1050 kg

Explanation:

The formula for kinetic energy is:

KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.

We can now substitute the values we know into this equation.

KE = 472 500 J and v = 30 m/s:

472 500 = 1/2 × m × 30²

Next, we can rearrange the equation to make m the subject and solve for m:

m = 472 500 ÷ (1/2 × 30²)

m = 472 500 ÷ 450

m = 1050 kg

Hope this helps!

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