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2 years ago

Example of first, second and third law of motion.

Physics

1 answer:

statuscvo [17]2 years ago

4 0

Examples of Newton's three law of motion.

First law of motion: A rocket being launched up in the atmosphere.

Second law of motion:while riding a bicycle, a bicycle acts as a mass and our legs pushing on the pedals of the bicycle is the force.

Third law of motion:when we jump off from the boat,the boat moves backward.

Hope,it will helpyouu!

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Ymorist [56]

Answer: I would say the answer is B.

Explanation: It looks and sounds the most correct.

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2 years ago

A 50n box is lifted 2 meters in 3 seconds.

anygoal [31]

F=G=50 N

L=F*d=50*2=100 J

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2 years ago

Reflected waves change their wavelength by ______ when reflected.

natita [175]

Answer:

i think its c

Explanation:

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2 years ago

A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =

Liula [17]

<span>A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.

4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?</span>

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2 years ago

A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag

VARVARA [1.3K]

Answer:

Induced emf, $\epsilon=9.79\times 10^7\ volts$

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, $\omega=3\ rev/s=18.84\ rad/s$

The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

$\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2$

$\epsilon=9.79\times 10^7\ volts$

So, the induced emf between the tip of a blade and the hub is $\epsilon=9.79\times 10^7\ volts$ . Hence, this is the required solution.

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2 years ago

Example of first, second and third law of motion.​

Example of first, second and third law of motion.