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11 months ago

11

Auto companies frequently test the safety of automobiles by putting themthrough crash tests to observe the integrity of the pass

enger compartment.If a 1000-kg car is sent toward a cement wall with a speed of 14 m/s and theimpact brings it to a stop in 8.00 x 10-2 s, with what average force is it,brought to rest?

1 answer:

e-lub [12.9K]11 months ago

4 0

Answer:

The average force that brings the car to rest is 175000N

Explanation:

The mass of the car, m = 1000 kg

Initial speed, u = 14 m/s

Final speed, v = 0 m/s (Since the car comes to a stop)

The time taken, t = 8 x 10^-2 s

The average force is calculated as:

$\begin{gathered} F=\frac{m(u-v)}{t} \\ F=\frac{1000(14-0)}{8\times10^{-2}} \\ F=\frac{14000}{0.08} \\ F=175000N \end{gathered}$

The average force that brings the car to rest is 175000N

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Answer:

The maximum frequency of revolution is 3.6 Hz.

Explanation:

Given that,

Mass = 8 kg

Distance = 400 mm

Tension = 800 N

We need to calculate the velocity

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v= velocity

r = radius of circle

Put the value into the formula

$800=\dfrac{8\times v^2}{200\times10^{-3}}$

$v^2=\sqrt{\dfrac{800\times200\times10^{-3}}{8}}$

$v=4.47\ m/s$

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$f=3.6\ Hz$

Hence, The maximum frequency of revolution is 3.6 Hz.

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Answer:

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Explanation:

In the first case there's an acceleration that modifies the direction of the movement.

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3 years ago

A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?

lapo4ka [179]

<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

$V_{f}^{2}-V_{o}^{2}=2.a.d$ (1)

Where:

$V_{f}$ is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

$V_{o}$ is the Initial Velocity, the value we want to find

$a$ is the constant acceleration of the car (the negative sign means the car is decelerating)

$d$ is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find $V_{o}$ :

$0-V_{o}^{2}=2(-5m/{s}^{2})(15m)$ (2)

$-V_{o}^{2}=-150{m}^{2}/{s}^{2}$ (3)

Multiplying by -1 on both sides of the equation:

$V_{o}^{2}=150{m}^{2}/{s}^{2}$ (4)

$V_{o}=\sqrt{150{m}^{2}/{s}^{2}}$ (5)

Finally:

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2 years ago

Which of the following involves an endothermic reaction?

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The answer would be c

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3 years ago

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When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

pantera1 [17]

Answer:

The period is $T = 0.700 \ s$

Explanation:

From the question we are told that

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The extension of the spring is $x = 12.0 \ cm = 0.12 \ m$

The spring constant for this is mathematically represented as

$k = \frac{F}{x}$

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$F = mg = 0.350 * 9.8$

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So

$k = \frac{3.43 }{ 0.12}$

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$T = 2 \pi \sqrt{\frac{m}{k} }$

substituting values

$T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }$

$T = 0.700 \ s$

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3 years ago