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Leokris [45]
2 years ago
5

The students on the right are applying a force of 100 N to the right. This is shown by the red arrow. The students on the left a

re applying a force of 100 N to the left. This is shown by the blue arrow. A student on the blue team loses balance and lets go of the rope. What is the most likely effect of this change?
A. The rope will move to the right, in the direction of the net force.

B. The rope will move to the left, in the direction opposite to the net force.

C. The rope will move to the left, in the direction of the net force.

D. The rope will move to the right, in the direction opposite to the net force.

Physics
2 answers:
kipiarov [429]2 years ago
6 0
The answer is D.
…………..
brilliants [131]2 years ago
4 0
I would say D because when you think about tug-of-war, when someone on one side lets go, the people on the other team are able to pull the rope further to their side. So the rope would move right to the side of the red team because someone on the blue team let go therefore, there is less balancing the rope on the blue teams side.
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7 0
3 years ago
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and ex
Ahat [919]

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, d = 1.8 \ m

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point D, the speakers are out of phase and so the path difference is $=\frac{\lambda}{2}$

Therefore,

$AD-BD = \frac{\lambda}{2}

$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$

$\lambda = 2 \times 0.4985$

$\lambda = 0.99714 \ m$

Thus the frequency is :

$f=\frac{v}{\lambda}$

$f=\frac{340}{0.99714}$

f=340.9744 Hz

3 0
2 years ago
HELPPPPP !!!!
Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0
2 years ago
A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0
3 years ago
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This is an electrical device that changes the voltage of electricity through the use of two set of coils
Fynjy0 [20]
Voltmeter is the right answer
3 0
3 years ago
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