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3 years ago

It takes 185 kj of work to accelerate a car from 23.0 m/s to 28.0 m/s. what is the car's mass?

1 answer:

6 0

<span>This question is based on conservation of energy as the work done would lead to change in kinetice energy of car change in KE = 1/2 mv(f)^2 - 1/2mv(i)^2 = 1/2m(v(f)^2-v(i)^2) where v(f) and v(i) are the final and initial speeds change in KE = 185kJ = 185,000J = 1/2 m((28m/s)^2-(23m/s)^2) 185,000=1/2 m(255m^2/s^2) solving for m m=1451kg</span>

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When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

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Question 30

stellarik [79]

Answer: $0.69\°$

Explanation:

The angular diameter $\delta$ of a spherical object is given by the following formula:

$\delta=2 sin^{-1}(\frac{d}{2D})$

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$d=16 m$ is the actual diameter

$D=1338 m$ is the distance to the spherical object

Hence:

$\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})$

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