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3 years ago

You drive a car for 2.0 h at 60 km/h, then for another 3.0 h at 85 km/h. What is your average velocity

Physics

1 answer:

Anon25 [30]3 years ago

8 0

Answer:

Explanation:

Average velocity is found in the total displacement experienced in the trip divided by the total time it took to make this trip. If, for the first 2.0 hours, the car travels at 60 km/hr, then in 2.0 hours the car can travel 120 km; if it then travels for 3.0 at 85 km/hr, it can travel 255 km in the same direction. The total time this took was 5.0 hours. So doing the math and rounding correctly by following the rules for the adding and subtracting of sig fig's:

$v=\frac{120+255}{5.0}=\frac{380}{5.0}=76\frac{km}{hr}$

If you do not round when you add, the average velocity is 75 km/hr

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If you have 12 atoms of hydrogen before a chemical reaction how many atoms of hydrogen will be president after the chemical reac

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2 years ago

A wave transfers: Water particles energy matter

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Answer:

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Explanation:

hope this helps u ....

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3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

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3 years ago