Ask question

Ask question

All categories

3 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

1 answer:

Alex73 [517]3 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

You might be interested in

The maximum force that a grocery bag can withstand without ripping is 250 n. Suppose that the bag is filled with 20 kg of grocer

zhannawk [14.2K]

Answer:

Groceries stay in the bag.

Explanation:

Given:

Maximum force = 250 N

Bag filled with = 20 kg

Lifted acceleration = $5.0\ m/s^2$

Solution:

We need to calculate the exerted force on the grocery bag by using Newton's second law.

$F = ma$

Where:

F = Exerted force on the object.

m = Mass of the object in kg

a = Acceleration of the object in $m/s^2$

Now, we substitute m = 20 kg and a = $5.0\ m/s^2$ in Newton's second law,

$F = 20\times 5.0$

$F = 100\ m/s^2$

Since, the exerted force on the bag is less than 250 N, the groceries will stay in the bag.

3 0

3 years ago

Gold-leaf electroscope uses

Kruka [31]

-identifies an electric charge
-it can identify its polarity (positive or negative) if you compare it to a charge that you already know
-can identify the magnitude of a charge (how big of a charge it is)

7 0

3 years ago

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0 -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 m $Vf^{2}$ = 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 m $Vi^{2}$ =0.5×200× $Vi^{2}$ =100 $Vi^{2}$ j

Put these values in eq 1, we get;

0-100 $Vi^{2}$ +3018.4-0=0

-100 $Vi^{2}$ =-3018.4

==> $Vi^{2}= \frac{3018.4}{100}$ = 30.184

==> Vi = $\sqrt{30.184} = 5.5 m.sec$

7 0

3 years ago

A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind

Marat540 [252]

Answer:

Part a)

$F_v = 4.28 N$

Part B)

$L = 1.02 m$

Part C)

$v = 1.25 m/s$

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

$Tcos\theta = mg$

$T sin\theta = F_v$

$\frac{F_v}{mg} = tan\theta$

$F_v = mg tan\theta$

$F_v = 1.2\times 9.81 (tan20)$

$F_v = 4.28 N$

Part B)

Here we can use energy theorem to find the distance that it will move

$-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2$

$(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2$

$(-3.5 + 2.54)L = - 0.98$

$L = 1.02 m$

Part C)

At terminal speed condition we know that

$F_v = mg$

$bv^2 = mg$

$2.5 v^2 = 3.9$

$v = 1.25 m/s$

7 0

2 years ago

Tim made tomato soup from scratch. When he cleaned up, he noticed that the aluminum pan was damaged. Analyze what happened to th

mr_godi [17]

Answer:

D) The acidic tomato juice reacted unfavorably with the aluminum pan

5 0

3 years ago

Read 2 more answers