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Gre4nikov [31]
3 years ago
11

what is the mass of an object that is experiencing a net force of 200N and an acceleration of 500m/s2​

Physics
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

F=200N

a=500m/s2​

Mass=?

Explanation:

F=ma

200=m*500

200/500=m

Mass=0.4kg

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When the volume of the gas is reduced, what change in property would be the most reasonable to expect?
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Answer:

Two possibities: Increase in pressure or decrease in temperature.

Explanation:

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The momentum of a 3000 kg truck is 6.36 x 104 kg·m/s. At what speed is the truck traveling? m/s
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Answer:

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Explanation

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P=6.36 x 10⁴kg·m/s.

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6 0
3 years ago
If gas in a cylinder is maintained at a constant temperature​ T, the pressure P is related to the volume V by a formula of the f
Mkey [24]

The given question is incomplete. The complete question is as follows.

If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form

P = \frac{nRT}{(V - nb)} - \frac{an^2}{V^2}, in which a, b, n, and R are constants. Find \frac{dP}{dV}.

Explanation:

We will use the quotient rule for each of the two terms on the right side as follows.

        P = \frac{nRT}{V - nb} - \frac{an^{2}}{V^{2}}

\frac{dP}{dV} = \frac{0(V - nb) - nRT(1)}{(V - nb)^{2}} - \frac{0(V)^{2} - an^{2}(2V)}{V^{4}}

            = \frac{-nRT}{(V - nb)^{2}} - \frac{-2an^{2}V}{V^{4}}

            = \frac{-nRT}{(V - nb)^{2}} + \frac{2an^{2}}{V^{3}}

            = \frac{2an^{2}}{V^{3}} - \frac{nRT}{(V - nb)^{2}}

   \frac{dP}{dV} = \frac{2an^{2}}{V^{3}} - \frac{nRT}{(V - nb)^{2}}

Thus, we can conclude that the value of \frac{dP}{dV} = \frac{2an^{2}}{V^{3}} - \frac{nRT}{(V - nb)^{2}}.

6 0
4 years ago
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