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Fittoniya [83]
2 years ago
14

1. A 700kg racecar slowed from 30m/s to 15 m/s. What was its change in Momentum? 2. If the car in the above problem took 5 secon

ds to slow down from 30m/s to 15 m/s. a. What is the impulse? b. What was the average force applied by the brakes?
Physics
1 answer:
s2008m [1.1K]2 years ago
6 0
1. The change in momentum follows the formula:

Change in Momentum = mΔv,
where m is the mass and v is the velocity

Change in momentum = (700 kg)(15 m/s - 30 m/s) =<em> -10,500 kg·m/s</em>

2. For this problem, the formula for impulse, J, is

J = FΔt

Before answering part a, we answer for part b first.

Part B. 
F = ma
where
a = Δv/t = (15 - 30)/5 = -1 m/s²
So,
F = (700 kg)(-1 m/s²) = <em>-700 N</em>

Part A.
J = (-700 N)(5 s) =<em> -3,500 N·s</em>
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Answer:

the molecular formula for the gas is NO₂

Explanation:

since it contains

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Oxygen = o → 69.55%

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Also we know that the proportion of oxygen over nitrogen  is

proportion of oxygen over nitrogen  = moles of oxygen / moles of nitrogen

since

moles = mass / molecular weight

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3 years ago
Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k1 and k2 and are connected via a thi
77julia77 [94]

Answer:

k1 + k2

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Spring 1 has spring constant k1

Spring 2 has spring constant k2

After being applied by the same force, it is clearly mentioned that spring are extended by the same amount i.e. extension of spring 1 is equal to extension of spring 2.

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Considering the general equation of spring forces (Hooke's Law) F = kx,

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2 years ago
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