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Contact [7]
3 years ago
9

By newton third law of motion, we know that if a rocket ship pushes down on the ground, the ground will push back up on the rock

et ship with twice as much force
Physics
2 answers:
Inessa [10]3 years ago
8 0
By Newton's third law of motion, we know that if a rocket ship pushes down on the ground, the ground will push back up on the rocket ship with twice as much force. ... Suppose an object has 20 N of tension force pulling it upward and 40 N of gravitational force pulling it downward.
Furkat [3]3 years ago
6 0

We don't know that at all. The 3rd law says that the REaction is opposite and EQUAL to the action.  We don't know where that "twice as much" comes from.

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A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a
Doss [256]

Answer:

Part(a): The frequency is \bf{0.2~Hz}.

Part(b): The speed of the wave is \bf{0.5~m/s}.

Explanation:

Given:

The distance between the crests of the wave, d = 2.5~m.

The time required for the wave to laps against the pier, t = 5.0~s

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is \lambda = 2.5~m.

Also, the time required for the wave for each laps is the time period of oscillation and it is given by T = 5.0~s.

Part(a):

The relation between the frequency and time period is given by

\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Substituting the value of T in equation (1), we have

\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz

Part(b):

The relation between the velocity of a wave to its frequency is given by

v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Substituting the value of \nu and \lambda in equation (2), we have

v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s

5 0
3 years ago
A man does 500 J of work pushing a car a distance of 2 m. How much force does he apply? Assume there is no friction.
Dmitry [639]

The correct answer is A. 250N

Work is a product of force and distance.

That is, work done=force×distance

Therefore substituting for the values in the question:

500J=force×2m

force= 500Nm/2m=250N

another unit for work done is Nm as force as the SI unit of force is newtons and distance in meters

6 0
3 years ago
Read 2 more answers
Proof that the kinetic energy of a moving body is half mv square​
Ivenika [448]

Answer:

1

Explanation:

4 0
2 years ago
if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put​
kari74 [83]

Answer:

300J

Explanation:

Work done = Force x the distance travelled in the direction of the force

=300 x 1

=300J

8 0
3 years ago
A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of
algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y =  A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0
3 years ago
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