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Murljashka [212]
4 years ago
13

..................................................................... how do i delete this

Physics
1 answer:
creativ13 [48]4 years ago
6 0

Answer:

LM.AO no clue....

Explanation:

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Tarzan swings back and forth on a long vine with a period of 7.27 s. how long is the vine?(unit=m)
lana [24]

Answer:

Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.

Explanation:

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4 years ago
Sound is a type of _____ wave.
saul85 [17]
Sound is an example of a mechanical wave. Mechanical waves are the kinds of waves that cannot be propagated without a medium. As such, these waves cannot travel through a vacuum, just like how sound cannot travel through space, since space is a vacuum.
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3 years ago
A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally holding a stopwatch, n
valkas [14]

Answer

given,

height of window = 4 m

time taken to travel = 1 s

acceleration due to gravity = 9.8 m/s²

s = ut + \dfrac{1}{2}at^2

u = \dfrac{s - \dfrac{1}{2}at^2}{t}

u = \dfrac{4 - \dfrac{1}{2}\times 9.8\times 1^2}{1}

u = -0.905 m/s

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now,

v² = u² + 2 a s

(-0.905)² = 0 + 2 × 9.8 ×s

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6 0
3 years ago
Choose the 200 kg refrigerator. Set the applied force to 400 N (to the right). Be sure friction is turned off.What is the net fo
White raven [17]
So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

We have an applied force of 400 N to the right, this means that:
<span>The magnitude of the net force is 400, directed to the right.</span>
3 0
3 years ago
Read 2 more answers
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
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