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polet [3.4K]
3 years ago
11

Which of the following is a real-life example of a rotation?

Physics
1 answer:
WITCHER [35]3 years ago
4 0

Answer:

a ceiling fan a ceiling fan

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A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change the volume to 43L?​
Licemer1 [7]

Answer:

roughly 2.2 atm

Explanation:

using Boyle's law (P1V1=P2V2),

73×1.3=43×x

x= 2.21 or 2.2 atm

8 0
3 years ago
What's an example sentence using thermal energy
xeze [42]
Friction is an example of thermal energy. As the tiny heat particles move, more warmness is made from the friction.
6 0
4 years ago
Find the net force and acceleration. 15 points. Will give brainliest!
gladu [14]

Answer:

\boxed{F_{net} = 28.7 \ N}

\boxed{a = 2.1 \ m/s^2}

Explanation:

<u><em>Finding the net force:</em></u>

<u><em>Firstly , we'll find force of Friction:</em></u>

F_{k} = (micro)_{k}mg

Where (micro)_{k} is the coefficient of friction and m = 13.6 kg

F_{k} = (0.16)(13.6)(9.8)\\

F_{k} = 21.32 \ N

<u><em>Now, Finding the net force:</em></u>

F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\

F_{net} = 28.7 \ N

<u><em>Finding Acceleration:</em></u>

a = \frac{F_{net}}{m}

a = \frac{28.7}{13.6}

a = 2.1 \ m/s^2

8 0
4 years ago
As a falling apple, approaches the ground, its potential energy
attashe74 [19]

Answer:

Decreases.

Explanation:

When an object with potential energy falls to the ground, its potential energy decreases and kinetic energy increases.

7 0
2 years ago
A 26-kg sled is on a snow-covered slope. The coefficients of friction between the sled’s runners and the snow are µs = 0.096 and µ
sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction \mu _s=0.096

coefficient of kinetic friction \mu _k=0.072

In order to move sled from rest we need to provide a force greater than static friction which is given by

f_s=\mu mg=0.096\times 26\times 9.8=24.46 N

After Moving Sled kinetic friction comes in to play which is less than static friction

f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0
3 years ago
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