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zhenek [66]
3 years ago
9

In this experiment, three different tuning forks were used. If we assume that a percent error of 5% is acceptable in determining

the speed of sound, what can we conclude about the speed of sound depend on the frequency of the tuning fork?
Physics
1 answer:
olya-2409 [2.1K]3 years ago
6 0
In the experiment, you had three tuning forks of frequency:
f₁ = 512 Hz, f₂ = 480 Hz and f₃ = 426 Hz.

Through the experiment you were able to calculate the speed of sound, finding:
v₁ = 340 m/s, v₂ = 342 m/s and v₃ = 341 m/s.

The difference in these measurements is within the 5% error, therefore you can infer that the speed of sound is independent of the frequency of the tuning forks.
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Graduated cylinders have numbers on the side that help you measure volume
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Consider a 24.0 V storage battery that can transfer, over the course of its useful lifetime, a total charge equivalent to 340000
xeze [42]

Answer:

Explanation:

Given that,

Potential difference of battery is

V = 24V

Total charge battery can transferred

q = 340,000C

Work done by the battery?

Work done is given as

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Where q is charge in Columbs

V is potential difference in Volts

Then, W = qV

W = 340,000×24

W = 8,160,000 J

Work done by the battery over it's useful life time is 8,160,000J

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3 years ago
How does the force of gravity exerted
givi [52]

The two forces of gravity are equal

Explanation:

We can answer this question by applying Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In this problem, we can identify the Sun as object A and the Earth as object B. This means that the force of gravity exerted by the Sun on the Earth is the action, while the force of gravity exerted by the Earth on the Sun is the reaction: according to Newton's third law, these two forces are equal and opposite.

Therefore, the two forces of gravity are equal in magnitude, which is given by:

F=\frac{GMm}{r^2}

where

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m is the mass of the Earth

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8 0
3 years ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
SSSSS [86.1K]

Answer:

{F_{tot} = -1.092*10^{-2}N

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

F =k\dfrac{q_1 q_2}{d^2}.

where d is the separation between the charges q_1 and q_2.

Now, in our case

q_1=-13.5nC=-13.5*10^{-9}C

q_2=-1.735nC=-1.735*10^{-9}C

q_3= 47*10^{-9}C

The separation between charges q_3 and q_1 is

d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m

Therefore, the force between them is

F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N

and it is directed in the negative x-direction.

The separation between charges q_2  and q_3 is

d_{23} =-1.240mm-0=-1.240*10^{-3}m

therefore, the force between them is

F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N

Therefore the total force on charge q_3 is

F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}

8 0
3 years ago
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