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lord [1]
3 years ago
9

2) what is altitude?

Physics
2 answers:
Ronch [10]3 years ago
5 0
2) b) The height above the surface of the earth

3) b) air pressure differences that make air rise or sink

4) a) TC (I'm not sure about this)

5) I don't know this one. Sorry :(

6) a) true

7) a) true

8) b) false (I'm not sure about this)

Sorry I couldn't answer all your questions, but I hope this helps!

lubasha [3.4K]3 years ago
5 0
8 a true ... just watch a UK weather forecast = wind and rain generally ...
not so at the Azores - high pressure tourist attraction ... yum yum
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An arrow strikes a target moving at 75 m/s and embeds itself 15 cm into the target. If the arrow stopped with constant accelerat
zhenek [66]

Answer:

Answer:

4 ms

Explanation:

initial velocity, u = 75 m/s

final velocity, v = 0

distance, s = 15 cm = 0.15 m

Let the acceleration is a and the time taken is t.

Use third equation of motion

v² = u² + 2 a s

0 = 75 x 75 - 2 a x 0.15

a = - 18750 m/s^2

Use first equation of motion

v = u + at

0 = 75 - 18750 x t

t = 4 x 10^-3 s

t = 4 ms

thus, the time taken is 4 ms.

Explanation:

5 0
3 years ago
Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R.
Dennis_Churaev [7]

Answer:

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

Explanation:

The rotational kinetic energy when the cylinder is with the rope is:

E_k=\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^2

where we used the fact that both rope and cylinder hast the same w. This E_k must conserve, that is, E_k must equal E_k when the rope leaves the cylinder. Hence, the final w is given by:

E_{k1}=E_{k2}\\\\\frac{1}{2}I_c\omega_0^2+\frac{1}{2}I_r\omega_0^{2}=\frac{1}{2}I_c\omega^2\\\\\omega=\sqrt{\omega_0^2(\frac{I_c+I_r}{I_c})} (1)

For Ic and Ir we can assume that the rope is a ring of the same radius of the cylinder. Then, we have:

I_c=\frac{1}{2}MR^2\\\\I_r=mR^2

Finally, by replacing in (1):

\omega=\sqrt{\omega_0^2(\frac{M+m}{M})}

hope this helps!!

7 0
2 years ago
The binding of acetylcholine to its receptor at the neuromuscular junction causes the opening of a
Rufina [12.5K]

Answer: opening of the nicotinic acetylcholine receptor channels.

Explanation:

Neuromuscular junction is a special junction formed between a motor neurone and a muscle fibre. The junction is fortified with nerves and receptors that helps in the transmission of signals from the motor neurone to the muscle fibre in order to bring about the desired voluntary movements through muscular contraction.

Nicotinic acetylcholine receptor are activated through the binding of acetylcholine at the neuromuscular junction. This action leads to influx of sodium ions to accomplish endplate potential.

7 0
2 years ago
why are we all cheating if we payed attention in what they trying to tell us we wouldn't have to cheat but we are not paying att
sweet-ann [11.9K]
I’ve always been failing since middle school. it’s bcs of quarantine that made me unmotivated. rn my grades are F’s D C and A . I should be paying attention but my phone just keeps me distracted lol.
7 0
3 years ago
Read 2 more answers
The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?
katovenus [111]
<h3><u>Question: </u></h3>

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

<h3><u>Answer:</u></h3>

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

<h3><u> Explanation: </u></h3>

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence F_{G} = F_{C}.

Gravitational force between Earth and Satellite: F_{G} = \frac{G \times M_e \times M_s}{R^2}

Centripetal force of Satellite :F_C = \frac{M_s \times V^2}{R}

Where G = Gravitational Constant

M_e= Mass of Earth

M_s= Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating  F_G = F_C, we get  

Speed of Satellite V =\frac{\sqrt{G \times M_e}}{R}

Thus the speed of satellite depends only on the mass of Earth.

6 0
3 years ago
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