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aliya0001 [1]
4 years ago
15

Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through e

xperimental data, the students determine that the mass of the object is distributed radially. They determine that the radius of the object as a function of its mass is given by the equation r = km², where k = 3.
Which of the following is a correct expression for the rotational inertia of the object?

(A) m3
(B) 1.8 m3
(C) 3.6 m3
(D) 6 m3
(E) 9 m3
Physics
1 answer:
Alex73 [517]4 years ago
5 0

Answer: I = 3.6 m3

(C)

Explanation:

moment of inertia for spherically shaped object around it's center is given as

I = (2/5) mr²

substituting the r = 3m²

I = (2/5)*(9) m3

I = 3.6 m3

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Answer:

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b. F = 42.6 N

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Explanation:

a.

First, we find the change in momentum of the bullets. For one bullet:

ΔP = m(Vf - Vi)

where,

ΔP = Change in Momentum = ?

m = mass of bullet = 5 x 10⁻³ kg

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Vi = Initial Speed = 0 m/s (Since bullets are initially at rest)

Therefore,

ΔP = (3 x 10⁻³ kg)(1110 m/s - 0 m/s)

ΔP = 3.33 N.s

For 151 bullets:

ΔP = (151)(3.33 N.s)

ΔP = 502.83 N.s

Now, dividing this by time interval, Δt = 11.8 s

ΔP/Δt = 502.83 N.s/ 11.8 s

<u>ΔP/Δt =  42.6 N</u>

<u></u>

b.

According to Newton's Second Law, the force is equal to rate of change of linear momentum:

Average Force = F = ΔP/Δt

<u>F = 42.6 N</u>

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c.

The pressure is given by:

Average Pressure = P = Average Force/Area

P = 42.6 N/ 3 x 10⁻⁴ m²

<u>P = 142042.4 Pa = 1.42 KPa</u>

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A solid of density 8000 kgm.. weighs 0.8 kgf in air. When it is completely
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Answer:<em> 0.5kg</em>

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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
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Answer:

F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

|F_T|=2\sqrt{34}k\frac{Q^2}{L}

\theta=tan^{-1}(\frac{5}{3})=59.03\°

Explanation:

I attached an image below with the scheme of the system:

The total force on the charge 2Q is the sum of the contribution of the forces between 2Q and the other charges:

F_T=F_Q+F_{3Q}+F_{4Q}\\\\F_T=k\frac{(Q)(2Q)}{R_1}\hat{i}+k\frac{(3Q)(2Q)}{R_2}\hat{j}+k\frac{(4Q)(2Q)}{R_3}[cos\theta \hat{i}+sin\theta \hat{j}]

the distances R1, R2 and R3, for a square arrangement is:

R1 = L

R2 = L

R3 = (√2)L

θ = 45°

F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[cos(45\°)\hat{i}+sin(45\°)\hat{j}]\\\\F_T=k\frac{2Q^2}{L}\hat{i}+k\frac{6Q^2}{L}\hat{j}+k\frac{8Q^2}{\sqrt{2}L}[\frac{\sqrt{2}}{2}\hat{i}+\frac{\sqrt{2}}{2}\hat{j}]\\\\F_T=6k\frac{Q^2}{L}\hat{i}+10k\frac{Q^2}{L}\hat{j}=2k\frac{Q^2}{L}[3\hat{i}+5\hat{j}]

and the magnitude is:

|F_T|=2k\frac{Q^2}{L}\sqrt{3^2+5^2}=2\sqrt{34}k\frac{Q^2}{L}

the direction is:

\theta=tan^{-1}(\frac{5}{3})=59.03\°

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