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dlinn [17]
3 years ago
14

Air resistance is a fluid type of friction. TRUE OR FALSE

Physics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

True

Explanation: Air resistance (also called drag) is a type of frictional force. Like all frictional forces, the force of air resistance always opposes the motion of an object. Usually, the air resistance force is not very strong.

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Followed by the previous question: presume that the electron performs a uniform circular motion around the hydrogen nucleus. Wha
Ksivusya [100]

Answer:

A_c=87.73*10^{21}m/s

Explanation:

From the question we are told that

r=5\times 10^{-11}

T=1.5 \times 10^{-16}

Generally the equation for velocity is mathematically given as

Velocity (v)=\frac{2 \pi r}{t}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

V=\frac{2 \pi (5*10^{-11})}{1.5*10^{-16}}

Generally the equation for Centripetal acceleration is mathematically given as

A_c=\frac{V^2}{r}

A_c=(\frac{20.944*10^5)}{r5*10^{-11}}

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8 0
3 years ago
Oxygen enters the blood through the.... I do not know the answer. Can someone please help me?!?!
frez [133]
Oxygen enters the blood Through your lungs
8 0
3 years ago
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Find equivalent resistance between A and B​
V125BC [204]

Answer:

In the picture

Explanation:

I hope that it's a clear solution and explanation, hope that helps.

8 0
3 years ago
Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

A_y=+15 m

\theta=32^{\circ}

We have to find the x-component of A and magnitude of A.

According to question

A_y=\mid A\mid cos\theta

Substitute the values then we get

15=\mid A\mid cos32

\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}

\mid A\mid=17.7m

tan\theta=\frac{perpendicular\;side}{Base}

tan32=\frac{A_x}{A_y}=\frac{A_x}{15}

0.62\times 15=A_x

A_x=9.3

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

6 0
3 years ago
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imagine riding a single-speed bicycle. why do you have to push harders on the pedals to start the bicycle moving than to keep it
Flura [38]
A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces. 
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