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3 years ago

What is the mass of an object that requires a force of 182 N to accelerate at a rate of 13 m/s?

Physics

1 answer:

Inga [223]3 years ago

6 0

Answer:

$m=14kg$

Explanation:

Hello.

In this case, since the force is defined in terms of the mass and acceleration by:

$F=m*a$

We can easily compute the mass by solving for it:

$m=\frac{F}{a}$

Whereas the force is 182 N (kg*m/s²) and the acceleration is 13 m/s², therefore, we obtain:

$m=\frac{182kg\frac{m}{s^2} }{13\frac{m}{s^2}}\\\\m=14kg$

Best regards.

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Use Coulomb’s law to derive the dimension for the permittivity of free space. <br><br><br><br>

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Answer:

Coulomb's law is:

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First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

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then the dimensions of e0 are:

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3 years ago

A car's momentum is p when it is traveling with velocity of v. if the velocity of that car doubled, what is the new momentum of

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3 years ago

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Answer:

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Find the electric flux passing through the 45° surface of an upright circular cone with a base diameter of 4 m if the electric f

alexgriva [62]

Answer:

electric flux = 177.625 N-m²/C

Explanation:

given data

angle = 45°

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electric field E = 20 N/C

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solution

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electric flux = E × A × cosθ ..............1

here A is area that is

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electric flux = 20 × 12.56 × cos45

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3 years ago

) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th

Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

Now, we will substitute 0 for both $z_{1}$ and $z_{2}$ , 0 for w, 334.9 kJ/kg for $h_{1}$ , 2726.5 kJ/kg for $h_{2}$ , 5 m/s for $V_{1}$ and 220 m/s for $V_{2}$ .

Putting the given values into the above formula as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

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q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

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