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Ymorist [56]
3 years ago
14

Donna stretched a spring to increase its length. Nearby in an electrical device, a positive charge attracted a negative charge.

Which statement is true about the spring and the negative charge?
A. They both experienced contact forces.
B. They both experienced non-contact forces.
C. The spring experienced a contact force and the charge experienced a non-contact force.
D. The spring experienced a non-contact force and the charge experienced a contact force.
Physics
1 answer:
timurjin [86]3 years ago
7 0

C. The spring experienced a contact force and the charge experienced a non-contact force

Explanation:

In physics, there are two types of forces:

  • Contact forces: contact forces are those where a contact between the objects involved is needed, in order to have the force. Examples of contact forces are the force of friction (the contact between the two surfaces is needed), or the elastic force in a spring (in fact, the spring needs to be physically in contact with something that stretches it)
  • Non-contact forces: non-contact forces are those where no contact is needed between the objects, so they can be exerted at a distance. Examples are the gravitational force (two masses exert a force on each other even from a distance) or the electric force (two charges exert a force on each other even from a distance).

Therefore, in this example:

  • The spring experiences a contact force
  • The two charges experience a non-contact force

So the correct answer is

C. The spring experienced a contact force and the charge experienced a non-contact force

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

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Answer:

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Explanation:

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Work done

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N

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viktelen [127]

Answer:

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Explanation:

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goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

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3 years ago
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

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