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krok68 [10]
2 years ago
8

Which term best describes the motion of the rope particles in relation to the motion of the rope wave shown in the photograph

Physics
2 answers:
Nady [450]2 years ago
3 0

Answer:

A: Perpendicular

Explanation:

The question is incomplete as it lacks the image of the rope wave motion.

However, as found on "estudyassistant", the options are;

A) Perpendicular

B) Circular

C) Longitudinal

D) Parallel

From all that, we can say that;

The rope's are moving simultaneously in the same pattern without touching each other.

This is therefore a mechanical wave being created with the motion having oscillations that are perpendicular to the direction of energy transfer of the ropes.

This is a definition of transverse waves because the rope particle motion is perpendicular to the wave motion.

Ivenika [448]2 years ago
3 0

Answer:

A: Perpendicular

Explanation:

Read above explanation.

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5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?
Degger [83]

Answer:

\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

<u>Finding </u><u>the</u><u> </u><u>velocity</u><u> </u>

\boxed{ \sf{velocity =  \frac{distance \: travelled \: }{time \: taken}}}

\dashrightarrow{ \sf{velocity =  \frac{200 \: m}{24 \: s} }}

\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}

Hope I helped!

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3 years ago
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denis23 [38]
Answer

Around 400 B.C.E, the Greek philosopher Democritus introduced the idea of the atom as the basic building block matter. Democritus though that atoms are tiny, uncuttable, solid particles that are surrounded by empty space and constantly moving at random.

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2 years ago
A satellite s is moving in an elliptical orbit around the earth . the mass of satellite is very small compared to mass of earth.
Contact [7]
The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>A satellite (s) is moving in an elliptical orbit around the earth has its angular momentum towards the earth changing in direction, but not in magnitude.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
8 0
3 years ago
Answer for a thanks
mariarad [96]
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7 0
3 years ago
Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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