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Troyanec [42]
3 years ago
10

Gases have an indefinite shape and volume

Physics
1 answer:
Sindrei [870]3 years ago
8 0

Yes that's true they do.

A sample of gas will change its shape and size, to take on the exact shape and size of whatever container you put it in.

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A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr
fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}

The values are

\displaystyle y_o=15\ m

\displaystyle v_o=14\ m/s

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

\displaystyle y=0

\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

Multiplying by 2

\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

3 0
2 years ago
Pam rubs a balloon on her head to generate a static charge. She holds the balloon up against a wall. Which of the following desc
cestrela7 [59]

Answer:

c

Explanation:

6 0
3 years ago
An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a
SCORPION-xisa [38]

Answer:

a_y=0.92m/s^2

Explanation:

To solve this problem we use the formula for accelerated motion:

y=y_0+v_{y0}t+\frac{a_yt^2}{2}

We will take the initial position as our reference (y_0=0m) and the downward direction as positive. Since the rock departs from rest we have:

y=\frac{a_yt^2}{2}

Which means our acceleration would be:

a_y=\frac{2y}{t^2}

Using our values:

a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2

5 0
3 years ago
What factors affect potential energy
jok3333 [9.3K]
Mass ,gravity and height
6 0
2 years ago
Read 2 more answers
. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.
cestrela7 [59]

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

8 0
3 years ago
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