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trapecia [35]
3 years ago
14

A ball is dropped from a height of 12 feet and returns to a height that is one-half of the height from which it fell. How far wi

ll the ball have traveled when it hits the ground for the fourth time?. . A. 12 feet. . B. 24 feet. . C. 3 feet. . D. 33 feet¥
Physics
2 answers:
almond37 [142]3 years ago
7 0

Answer: (D) 33 feet

First consider that the height is 12 feet and the ball is dropped from that height and the ball bounces back up to half the height that is 6 feet, then second time when the ball is dropped, according to the given question, the height becomes 6 feet and when the ball returns back then it returns back to half of the height that is 3 feet, following the similar path, we can write bounces one at a time:

1.  Ball is dropped from 12 feet and bounces back to 6 feet.

2. Ball is dropped from 6 feet and bounces back to 3 feet.

3. Ball is dropped from 3 feet and bounces back to 1.5 feet.

4. Ball is dropped from 1.5 feet and reaches the ground.

Adding the total distance traveled,we get

12+6+6+3+3+1.5+1.5=33 feet.

Liula [17]3 years ago
6 0
The best thing to do in order to calculate the distance of the ball taht would have traveled when it hits the ground for the fourth time is to list the height everytime it bounces. We calculate as follows:

<span>12+6+6+3+3+1.5+1.5 = 33 feet</span>
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A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M a
Yanka [14]

Answer:

a. F = Qs/2ε₀[1 - z/√(z² + R²)] b.  h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

Explanation:

a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?

The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by

E = s/2ε₀[1 - z/√(z² + R²)]

So, the net force on the small plastic sphere of mass M and charge Q is

F = QE

F = Qs/2ε₀[1 - z/√(z² + R²)]

b. At what height h does the sphere hover?

The sphere hovers at height z = h when the electric force equals the weight of the sphere.

So, F = mg

Qs/2ε₀[1 - z/√(z² + R²)] = mg

when z = h, we have

Qs/2ε₀[1 - h/√(h² + R²)] = mg

[1 - h/√(h² + R²)] = 2mgε₀/Qs

h/√(h² + R²) = 1 - 2mgε₀/Qs

squaring both sides, we have

[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²

h²/(h² + R²) = (1 - 2mgε₀/Qs)²

cross-multiplying, we have

h² = (1 - 2mgε₀/Qs)²(h² + R²)

expanding the bracket, we have

h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²

collecting like terms, we have

h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²

Factorizing, we have

[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²

So, h² =  (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]

taking square-root of both sides, we have

√h² =  √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]

h =  (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]

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3 years ago
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Of these options the one which best answers the question is 'Use whatever professionals claim to use.'  
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3 years ago
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ArbitrLikvidat [17]

Answer:

Explanation:

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8 0
3 years ago
Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.
Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

f_n = nf_1

where

n is the number of the harmonic

f_1 is the fundamental frequency

Here we have:

f_3=512 Hz is the frequency of the 3rd harmonic

So the fundamental frequency is

f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz

And so, the frequency of the 2nd harmonic is:

f_2=2f_1=2(170.7)=341.3 Hz

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