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igor_vitrenko [27]
3 years ago
8

If the graph shown is a position-time graph of an object moving at constant velocity, what is the velocity of the object? Assume

that the units of position are meters and the units of time are seconds.

Physics
1 answer:
belka [17]3 years ago
5 0

Answer:

C.) v = 50 m/s

Explanation:

The relationship between position vs. time graph and velocity is that the derivative (slope) of the position vs. time graph gives you velocity. In other words, find the slope to get the velocity.

Select two arbitrary points. I'll choose (3s, 225m) and (0s, 75m).

Now use the slope equation:

v = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} =\frac{225 - 75}{3-0} = 50 m/s

v = 50 m/s

Hopes this helps!

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- As you ride in a car, both you and the car are moving
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c. Your body is at rest but its inertia puts it in motion

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Physicists often measure the momentum of subatomic particles moving near the speed of light in units of MeV/c, where c is the sp
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Answer:

kg m/s

Explanation:

e = Charge = C

V = Voltage = \dfrac{N}{C}m

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Momentum is given by

\dfrac{MeV}{c}=\dfrac{e\times V}{c}\\\Rightarrow \dfrac{MeV}{c}=\dfrac{C\times \dfrac{N}{C}\times m}{m/s}\\\Rightarrow \dfrac{MeV}{c}=Ns\\\Rightarrow \dfrac{MeV}{c}=kg\times \dfrac{m}{s}\times s\\\Rightarrow \dfrac{MeV}{c}=kg\cdot m/s

The unit of MeV/c in SI fundamental units is kg m/s

5 0
3 years ago
The speed for a car that went a distance of 125 miles in 2 hours time is ______. Question 2 options: 250 meters/sec 62.5 miles/h
Goryan [66]

Answer:

62.5 miles per hour

Explanation:

Speed = Distance travelled / Time taken

Speed = 125/2 = 62.5

You derive the units of the speed...by using the speed formula....,

Speed = Distance/Time

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Hence the units for the speed = miles/hour

5 0
3 years ago
A 1,250 W electric motor is connected to a 220 Vrms, 60 Hz source. The power factor is lagging by 0.65. To correct the pf to 0.9
Black_prince [1.1K]

Answer:

C = 46.891 \mu F

Explanation:

Given data:

v = 220 rms

power factor = 0.65

P = 1250 W

New power factor is 0.9 lag

we knwo that

s = \frac{P}{P.F} < COS^{-1} 0.65

S = \frac{1250}{0.65} < 49.45

s = 1923.09 < 49.65^o

s = [1250 + 1461 j] vA

P.F new = cos [tan^{-1} \frac{Q_{new}}{P}]

solving for Q_{new}

Q_{new} = P tan [cos^{-1} P.F new]

Q_{new} = 1250 [tan[cos^{-1}0.9]]

Q_{new} = 605.40 VARS

Q_C = Q - Q_{new}

Q_C = 1461 - 605.4 = 855.6 vars

Q_C = \frac[v_{rms}^2}{xc} =v_{rms}^2 \omega C

C = \frac{Q_C}{ v_{rms}^2 \omega}

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C = 46.891 \mu F

6 0
4 years ago
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