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gulaghasi [49]
3 years ago
12

A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s/s. How long does it take

the car to come to a stop? Use the GUESS method. Please help me asap.
Physics
1 answer:
Maksim231197 [3]3 years ago
3 0
For this, we will use the formula:

<span>a = [V₀⁻V₁]÷t

</span>Where:

a - Acceleration
<span>V₀ - Final Velocity
</span>V₁ - Initial Velocity
<span>t - Time

</span>Given:
a = -2.0 m/s²
V₁ = 15.0 m/s
V₀ = 0 m/s ; because the car stopped

Find:
t = ?

Solution:

a = [V₀⁻V₁]÷t

First, we should derive the formula to the element that we are looking for (which is time).

t = [V₀⁻V₁]÷a
t = [0 m/s - 15.0 m/s]÷(-2.0 m/s²) ; substituting the numbers
t = -15.0 m/s ÷ -2.0m/s²
t = 7.5 seconds ; cancelling the removable units thus resulting only to the time unit s(seconds)

It took the car 7.5 seconds to stop
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A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
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Answer:

Explanation:

Given

Weight of person W=652\ N

At highest point Magnitude of the normal force F=585\ N

net force at highest point

F_{net}=W+F_c

where F_c= centripetal force

F_c=\dfrac{mv^2}{r}

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585=652+F_c

F_c=-67\ N

Negative sign shows force is in upward direction

At bottom point centripetal force is towards the bottom

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Read 2 more answers
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
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