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Natalka [10]
3 years ago
6

What is the force of friction between the dragsters tires and the track if the dragster mass is 300kg and the coefficient of the

friction force is 0.8
Physics
1 answer:
Nuetrik [128]3 years ago
7 0

Ff=μN=μmg=0.8*300*10=2400 N

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Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =
Ann [662]

Answer:

change in relative vorticity  0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C  per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g /  potential temperature ) × (potential vertical temperature) × exp^1/2    ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61  / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

so

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

8 0
3 years ago
A graph here shows the displacement of two walkers over time. Waliper 1 is graphed in red while walker 2 is graphed
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