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OLEGan [10]
3 years ago
9

An equation for the motion of an object is given as 2vt-4x=Bt^2 where B is a constant. The variable v indicates velocity, in met

ers per second; t indicates time, in seconds; and x indicates displacement, in meters. What is the unit of measure for B.?
Physics
2 answers:
Jlenok [28]3 years ago
8 0

Answer:

acceleration m/s²

Explanation:

The units on the left side of the equation are units of length. The units on the right side of the equation must be units of length also. Acceleration times time squared gives length.

Anestetic [448]3 years ago
5 0

Answer:

unit of B must be m/s/s

Explanation:

when two or more physical quantities added or subtracted then the unit of each part must be same

so here we have

2vt - 4 x = B t^2

so we can say that SI unit of each term must be equal to other term here

so for the first term

2 v t = (\frac{m}{s})(s) = meter

for next term we have

4 x = meter

so last term must have same unit

B t^2 = meter = B( sec^2)

so unit of B must be m/s/s

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U = 56.86 [J]

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U = 0.5*k*(x)^2

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A gas is in a sealed container.Part ABy what factor does the gas temperature change if the volume is doubled and the pressure is
Goryan [66]

A) The temperature increases by a factor 6

We can use the ideal gas equation:

pV=nRT

where

p is the pressure

V is the volume

n is the number of moles

R is the gas constant

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We can also rewrite it as

\frac{pV}{T}=nR

The gas is in a sealed container - this means the amount of gas is fixed, so n is constant. Since R is constant too, the term on the right in the equation is constant. So we can rewrite the equation as:

\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}

where in this problem we have:

V_2 = 2V_1 (the volume is doubled)

p_2 = 3 p_1 (the pressure is tripled)

re-arranging the equation, we find the change in temperature:

\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(2 V_1)}{p_1 V_1}=6

so, the temperature increases by a factor 6.

B) The temperature increases by a factor 1.5.

We can use again the same equation:

\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}

Where in this case:

V_2 = \frac{V_1}{2} (the volume is halved)

p_2 = 3 p_1 (the pressure is tripled)

So, we can find the change in temperature:

\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(\frac{V_1}{2})}{p_1 V_1}=\frac{3}{2}=1.5

So, the temperature increases by 1.5 times.

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3 years ago
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