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3 years ago

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A crash test dummy with a mass of 55kg is placed into a car traveling at 40m/s. During the first trial the dummy does not wear a

seatbelt and comes to rest in a time of 0.2 seconds. In the second trial, the dummy wears a seatbelt and comes to rest in a time of 0.5 seconds. Calculate the force experienced by the dummy with and without a seatbelt.

1 answer:

s2008m [1.1K]3 years ago

4 0

a. The force experienced by the dummy with a seatbelt is 11,000 Newton.

b. The force experienced by the dummy without a seatbelt is 4,400 Newton.

<u>Given the following data:</u>

Mass = 55 kg

Initial velocity = 0 m/s

Final velocity = 40 m/s

Time A = 0.2 seconds

Time B = 0.5 seconds

a. To calculate the force experienced by the dummy with a seatbelt:

<h3>Newton's Second Law of Motion</h3>

In order to solve for the force acting on the dummy, we would apply Newton's Second Law of Motion.

<u>Note:</u> The acceleration of an object is equal to the rate of change in velocity with respect to time.

Mathematically, Newton's Second Law of Motion is given by this formula;

$F = \frac{M(v\;-\;u)}{t}$

Substituting the given parameters into the formula, we have;

$F = \frac{55(40\;-\;0)}{0.2}\\\\F = \frac{55\times 40}{0.2}\\\\F = \frac{2200}{0.2}$

Force = 11,000 Newton.

b. To calculate the force experienced by the dummy without a seatbelt:

$F = \frac{55(40\;-\;0)}{0.5}\\\\F = \frac{55\times 40}{0.5}\\\\F = \frac{2200}{0.5}$

Force = 4,400 Newton.

Read more on force here: brainly.com/question/1121817

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olchik [2.2K]

Answer:

(A) the angular acceleration of the blades is 13.33 m/s.

Explanation:

Given;

moment of inertia of a blade, I = 0.2 kgm²

net torque exerted on fan blades, ∑τ = 8Nm

Torque is given as product of moment of inertia and angular acceleration;

τ = Iα

where;

α is the angular acceleration

Since there are three blades of the ceiling fan, the net torque is given as;

∑τ = (3I)α

∑τ = 3Iα

α = ∑τ / 3I

α = (8) / (3 x 0.2)

α = 13.33 m/s

Therefore, the angular acceleration of the blades is 13.33 m/s.

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3 years ago

You can enter compound units that are combinations of other units that are multiplied together.Torque can be calculated by multi

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Answer:

3.0 x 10¹ Nm

Explanation:

Torque = F x r

Where F is force applied and r is perpendicular distance from pivot point . r

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Here F = 15 N and r = 2.0 m

Torque

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3 years ago

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

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Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

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3 years ago

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Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

or

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.

Solving for $v$ we get:

$v \geq \sqrt{\dfrac{2GM}{R} }$

putting in numerical values gives

$v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }$

$\boxed{v\geq 5739.5m/s}$

in kilometers this is

$v\geq5.7395m/s.$

Hence, the minimum speed required is 5.7395km/s.

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3 years ago

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NeX [460]

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

$W =F*d$

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

$W = 294*5\\W = 1470 [J]\\$

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

$efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%$

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3 years ago