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jek_recluse [69]
3 years ago
8

If you push on the floor with 5 N for 15 seconds, how much work is done?

Physics
2 answers:
Maslowich3 years ago
4 0

Answer: It would be 75

Explanation: If your pushing on the floor at a rate of 5 for 15 seconds you would multiply the numbers and get 75

Schach [20]3 years ago
4 0
15*5=75
The amount of work that is done is 75.
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Which is a characteristic of projectile motion?
SVETLANKA909090 [29]
<h2>Right answer: It follows a curved path </h2>

The movement of a projectile is a movement in two dimensions (forming a curved path: a parabola shape) with <u>constant acceleration. </u>

<u> </u>

A projectile is any body or object that is thrown or projected by means of some force and continues in motion by its own inertia. This means the only force that acts on it while in motion is <u>the acceleration of gravity</u> (in this case we are on Earth, so the gravity value is 9.8\frac{m}{s^{2}}).  

Where gravity influences the <u>vertical movement</u> of the projectile, while <u>the horizontal movement</u> of the projectile is the result of the tendency of any object to remain in motion at a constant speed (according to Newton's 1st law of motion sometimes called Law of Inertia).

The other options are <u>incorrect</u> because are <u>false</u>:

-The forward motion negates air resistance: There is always at least a small percent of air resistance, as long as that movement is done on Earth.  

-It has variable acceleration: In projectile motion acceleration is constant (gravity acceleration) .

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5 0
3 years ago
Read 2 more answers
Ishani and John now try a problem involving a charging capacitor. An uncharged capacitor with C = 6.81 μF and a resistor with R
DENIUS [597]

Answer:

Q=81.72\times10^{-6}C

I=2.1\times10^{-5}A

Explanation:

The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):

Q=CV=(6.81\times10^{-6}F)(12V)=81.72\times10^{-6}C

The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):

I=\frac{V}{R}=\frac{12V}{5.8\times10^{5}\Omega}=2.1\times10^{-5}A

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The answer is going to be B
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