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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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Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
5. A family of ducks is swimming in a pond at a speed of 3 m/s when a gust of wind hits them. By the time they reach the other s
ella [17]

Answer:

The time taken by the duck to cross the lake is, t= 4 s

Explanation:

Given data,

The initial speed of the ducks, u = 3 m/s

The final speed of the ducks, v = 7 m/s

The acceleration of the duck, a = 1 m/s²

The formula for the acceleration is,

                               a = (v - u) / t

∴                               t = (v - u) / a

Substituting the given values in the above equation,

                                t = (7 - 3) / 1

                                  = 4 s

Hence, the time taken by the duck to cross the lake is, t= 4 s

6 0
3 years ago
According to the periodic table, which of the following elements has five energy levels
Alex Ar [27]
<span>Antimony I am pretty sure is one. </span>
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A piston/cylinder contains 2 kg of water at 20◦C with a volume of 0.1 m3. By mistake someone locks the piston, preventing it fro
morpeh [17]

Answer:

Hi

Final temperature = 250.11 °C

Final volume = 0,1 m3.

Process work = 0

Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

v=0.05 m3/kg

v1=0.057061 m3/kg

T1=242.56°C

v2=0.049779 m3/kg

T2=250.35°C

T=\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C

The process work is zero because there is no change in volume during heating:

W=PxΔv=Px0=0

where

W=process work

P=pressure

Δv=change of volume, is zero because the piston was blocked so the volume remains constant.

7 0
3 years ago
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Zigmanuir [339]

Answer:

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Explanation:

Hope this helped!!

8 0
2 years ago
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