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Delicious77 [7]
3 years ago
13

A force of 350n Acts on an object of mass 17.5 kg what acceleration does it produce​

Physics
1 answer:
UNO [17]3 years ago
3 0

Answer:

20 m/s/s

Explanation:

F=ma, 350=17.5 * a, a=20 m/s/s

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If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob
Natalija [7]

Answer:

Explanation:

Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:

T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m

By using the y=A\sin(\omega t)  \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}

The mean position is the position when <em>y</em> = 0, so:

\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi rad/s

and v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi} in centimeters (cm).

4 0
2 years ago
A positive rod is touched to a neutral sphere. There is a bar with positively charged particles attached to a ball with positive
solong [7]

Answer: A is correct

It is showing the direction that protons move in, making the sphere negative

Explanation:

From the figure, since the charges on the rod are completely positive, it shows that the protons are transferring from the sphere to the rod leaving the rod negatively charged.

Therefore, It is showing the direction that protons move in, making the sphere negative.

4 0
3 years ago
Willingness to take turns is one way we can express our attitudes in
Ilia_Sergeevich [38]
<span>The correct option is A, that is, non-verbal communication. Non-verbal communication is defined as the kind of communication that are passed across without using words. Examples of this are: shaking of hands, hugging, back-patting, facial expression, eye-contact and every form of communication that are make use of without the usage of words.</span>
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3 years ago
Hurricane sandy produced some of the greatest destruction along the new jersey coast in communities situated along narrow strips
FrozenT [24]

The correct answer to fill in the blank would be:

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What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at
pantera1 [17]

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

6 0
3 years ago
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