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Artemon [7]
4 years ago
13

A solid spherical ball and a hollow spherical ball made out of the same material are released from rest at the top of a ramp. Th

ey roll down the ramp without slipping to the bottom. show answer Incorrect Answer 50% Part (a) On what quantities does the speed of each ball at the bottom of the ramp depend? Select all that apply. Radius of the ball Distribution of mass within the ball Mass of the ball Height of the ramp Grade Summary Deductions 4% Potential 96% Submissions Attempts remaining: 2 (4% per attempt) detailed view 1 4%
Physics
1 answer:
ziro4ka [17]4 years ago
7 0

Answer:

1 ) Distribution of mass within the ball

2 ) Height of the ramp

Explanation:

Acceleration of a rolling body down an inclined plane is given by the following formula

a = g sinθ / ( 1 + k² / R² )

k is radius of gyration , R is radius of the spherical object ,

when acceleration is more , velocity will also be more .

for objects in which masses are lying in the periphery like in hollow sphere , the value of k²/R² will be high so denominator of the expression will be high so acceleration will be less , hence velocity on reaching the bottom will be less.

On mass of the ball , velocity will not depend .

If height is increased , ball will have acceleration for greater time so velocity will be high.

On radius it will not depend because , radius r and k increases proportionately.

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The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard
gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

|\Delta E_{ij}| = |E_i-E_j|

Our states are given by

E_1 = -11.7eV

E_2 = -4.2eV

E_3 = -3.3eV

In this way the energy balance for the states would be given by,

|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\

|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV

|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0
4 years ago
A block with a mass of 31.8 kg is pushed on a frictionless
OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

W = F.d.\cos\theta

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

W = F.d

Given:

F = 75 N

m = 31.8 kg

Final velocity v_f = 15.3 \frac{m}{s}

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\

Now we can determine the displacement from the following formula:

d = \frac{1}{2}a^2+v_0t+d_0

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

d = \frac{1}{2}at^2\\

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m

and the corresponding work:

W = F\cdot d=75 N\cdot 48.94 m =3670.5J


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3 years ago
The brightness of a star as seen from Earth is referred to as brightness.
jolli1 [7]

Technically it is referred to as <em>star brightness</em> but yes you have the general idea.

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4 years ago
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You are moving into an apartment and take the elevator to the 6th floor. Suppose your weight is 660 N and that of your belonging
Ivan

Answer:

Explanation:

Total weight

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660+1100=1760N.

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Using newton law of motion

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Then, N=W

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W=26,752J

W=26.752KJ

b. Work done on me alone is still need to go through the same process but will remove the weight of the belonging

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W=10,032J

W=10.032KJ

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