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olchik [2.2K]
3 years ago
9

What is the car's average velocity (in m/s) in the interval between t = 0.5 s to t = 2 s?

Physics
1 answer:
irakobra [83]3 years ago
4 0

Answer:

1.0\; \rm m \cdot s^{-1}.

Explanation:

Consider a time period of duration \Delta t. Let change in the position of an object during that period of time be denoted as \Delta x. The average velocity of that object during that period would be:

\displaystyle \bar{v} = \frac{\Delta x}{\Delta t}.

For the toy car in this question, the time interval has a duration of 2.0 \; \rm s - 0.5\; \rm s = 1.5\; \rm s. That is: \Delta t = 1.5\; \rm s. (One decimal place, two significant figures.)

On the other hand, what would be the change in the position of this toy car during that 1.5\; \rm s?

Note, that from readings on the snapshot in the diagram:

  • The position of the toy car was 0.1\; \rm m at t = 0.5\; \rm s (the beginning of this 1.5-second time period.)
  • The position of the toy car was 1.6\; \rm m at t = 2.0\; \rm s (the end of this 1.5-second time period.)

Therefore, the change to the position of this toy car over that time period would be \Delta x = 1.6\; \rm m - 0.1\; \rm m = 1.5\; \rm m. (One decimal place, two significant figures.)

The average velocity of this car over this period of time would thus be:

\displaystyle \bar{v} = \frac{\Delta x}{\Delta t} = \frac{1.5\; \rm m}{1.5\; \rm s} = 1.0\; \rm m \cdot s^{-1}. (Two significant figures.)

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A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
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Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

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3 years ago
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3 years ago
A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of
Alenkinab [10]

Answer:

T=4.24 N.m

Explanation:

Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

F=0.5 kg* 9.8 m/s^{2}= 4.9 N

and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus

so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N

and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m

anothe way to do it is,

T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result

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