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igor_vitrenko [27]

3 years ago

13

A 30 N block is being pulled along a horizontal surface with an acceleration of 6 m/s2 by a rope. The coefficient of kinetic fri

ction between the block and the surface is 0.5. What is the force applied by the rope?

1 answer:

nata0808 [166]3 years ago

6 0

Answer:

33N

Explanation:

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A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charl

Bond [772]

Answer:

Boyle's Law

$\therefore P_1.V_1=P_2.V_2$

Explanation:

Given that:

<u><em>initially:</em></u>

pressure of gas, $= P_1$

volume of gas, $= V_1$

<em><u>finally:</u></em>

pressure of gas, $= P_2$

volume of gas, $= V_2$

<u>To solve for final volume</u> $V_2$

<em>According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.</em>

<em>According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.</em>

But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.

Here nothing is said about the temperature, so we consider the Boyle's Law which states that <em>at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.</em>

Mathematically:

$P_1\propto \frac{1}{V_1}$

$\Rightarrow P_1.V_1=k\ \rm(constant)$

$\therefore P_1.V_1=P_2.V_2$

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A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical

loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

PE =0.92414J

KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

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