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olga nikolaevna [1]
3 years ago
5

A 140 g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.537 m. The ball makes 2.27 revolut

ions in a second. What is its centripetal acceleration?
Physics
1 answer:
klio [65]3 years ago
6 0

Answer:

109.13 rad/s^2

Explanation:

m = 140 g = 1.4 kg, r = 0.537 m, f = 2.27 rps

The centripetal acceleration is given by

a = r ω^2

a = r x (2 π f)^2

a = 0.537 x ( 2 x 3.14 x 2.27)^2

a = 109.13 rad/s^2

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8 0
2 years ago
An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in bet
Vera_Pavlovna [14]

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

4 0
3 years ago
A woman pushes an oak chest across an oak floor at a constant speed of 0.450 m/s. The chest has a mass of 40.0 kg, and the coeff
natima [27]

Answer:

A. 243 N

Explanation:

Friction is the force that opposes the relative motion between systems that are in contact.

This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.

First find the normal force which is the force that would point directly upwards to support weight of the block.

Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.

Given m=40 kg and g=9.80 m/s²

N force=40×9.80 =392N

Then find the force of friction which is given by the formula;

<em>F=μN where μ is friction coefficient for the  oak chest  and N is the normal force on the chest</em>

Given <em>μ</em>=0.620  and N force = 392 N then it will be;

F=0.620× 392 =243.04 N

Answer : 243 N

6 0
2 years ago
Help me besties ok thx
Pachacha [2.7K]

Answer:

1 \\ 0.5 \times 500 \times 20 \times 20 \\  \ 2 \\ 0.5 \times 0.6 \times 100 \times 100 \\ 3 \\ 1000 \times 0.5 \times 10 \times 10 \\ 1000 \times 0.5 \times 20 \times 20 \\ 1000 \times 0.5 \times 30 \times 30 \\ 4 \\ 0.5 \times 40000 \times 30 \times 30 \\ 5 \\  \sqrt{ \frac{2}{0.1 \times 0.5} }

7 0
2 years ago
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