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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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4 0
3 years ago
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Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 542 km above the earth's sur
Evgen [1.6K]
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite 
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g' 
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h): 
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)] 
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec 
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
6 0
3 years ago
A small 18 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe's kinetic energy?
In-s [12.5K]

Kinetic energy = (1/2) (mass) (speed²).

A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.

A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.

They're both correct.

8 0
3 years ago
PLEASEEEEEE HELP
emmainna [20.7K]

Answer:

8.89 m/s² west

Explanation:

Assume east is +x.  Given:

v₀ = 120 m/s

v = 0 m/s

t = 13.5 s

Find: a

v = at + v₀

0 m/s = a (13.5 s) + 120 m/s

a = -8.89 m/s²

a = 8.89 m/s² west

8 0
3 years ago
Heights of men on a baseball team have a​ bell-shaped distribution with a mean of 166 cm 166 cm and a standard deviation of 5 cm
kaheart [24]

Answer:

95 %

99.7 %

Explanation:

\mu = 166 cm = Mean

\sigma = 5 cm =  Standard deviation

a) 156 cm and 176 cm

166-5\times 2=156

166+5\times 2=176

From the empirical rule 95% of all values are within 2 standard deviation of the mean, so about 95% of men are between 156 cm and 176 cm.

b) 151 cm and 181 cm

166-5\times 3 =151

166+5\times 3=181

The empirical rule tells us that about 99.7% of all values are within 3 standard deviations of the mean, so about 99.7% of men are between 151 cm and 181 cm.

3 0
3 years ago
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