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Amiraneli [1.4K]
3 years ago
5

If tony uses a force of 1525 N to drive his car a distance of 381 m, how much work is done on his car?

Physics
2 answers:
Ipatiy [6.2K]3 years ago
7 0

Answer:

Explanation:

Yes i do, gud morning

shepuryov [24]3 years ago
3 0

Hello!!

For calculate the work let's applicate the formula:

\boxed{W=Fd}

\textbf{Being:}

\sqrt{} W = Work = ?

\sqrt{} F = Force = 1525 N

\sqrt{} d = Distance = 381 m

⇒ \text{Then let's \textbf{replace it according} we information:}

W = 1525 N * 381 m

⇒ \text{Let's resolve it: }

W = 581025 J

\textbf{Result:}\\\text{The work is \textbf{581025 Joules}}

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You are asked to give the answer in <span>g/cm3. So without knowing any single formulae you can just divide grams by cm3. 

</span>\frac{99 grams}{22 cm^{3} }

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7 0
3 years ago
The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

8 0
3 years ago
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d. Relative humidity increases.

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\phi = \frac{\omega \cdot P}{(0.622+\omega)\cdot P_{sat}}

When temperature decreases, the saturation pressure decreases also and, consequently, relative humidity increases. Therefore, the right answer is option D.

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3 years ago
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Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

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