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snow_lady [41]
3 years ago
14

A bowling ball of mass m = 1.7 kg drops from a height h = 14.2 m. A semi-circular tube of radius r = 6.2 m rest centered on a sc

ale. Write an expression for the reading of the scale when the bowling ball is at its lowest point, in terms of the variables in the problem statement.
Physics
1 answer:
marshall27 [118]3 years ago
8 0

Answer:

W_net = mg + 2mgh/r

Explanation:

The forces applied in this motion of the bowling ball are both gravitational and centripetal forces.

Now, gravitational force is; F_g = mg

While centripetal force is; F_c = mv²/r

Since we want to express the net force in terms of the variables in the statement and we are not given "v", let's find an expression of v with the variables given.

Now, from Newton's equation of motion, at initial velocity of 0, v² = 2gh.

Thus;

F_c = 2mgh/r

Where;

m is ball mass

r is tube radius

h is fall height

Thus, the net force will be;

F_net = F_g + F_c

Now, Net force would be equal to the net weight that will be read on the scale.

Thus;

W_net = F_net = F_g + F_c

W_net = mg + 2mgh/r

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A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?
Neko [114]

Answer: a=5 m/s^2

Explanation:

The acceleration of an object can be calculated by using Newton's second law:

F=ma

where

F is the net force applied on the object

m is the mass of the object

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In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

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5 0
3 years ago
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While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.
lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

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B is the answer, I’m really good at this subject
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