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3 years ago

Which is the example of circular motion??

Physics

2 answers:

AnnyKZ [126]3 years ago

6 0

ferris whell

none of them

Diano4ka-milaya [45]3 years ago

3 0

Explanation:

<h2>1) Ferris Wheel</h2>

this is your answer

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Hiii please help i’ll give brainliest if you give a correct answer please thanks!

lakkis [162]

Answer: the first one

Explanation: good luck!

8 0

3 years ago

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A person is standing on a spring bathroom scale on the floor of an elevator which is moving up and slowing down at the rate of 2

Alinara [238K]

Answer:

923.44N

Explanation:

Obviously Fn cannot equal Fg since the elevator is moving up

Where Fn is the force read on the scale

Fg is gravitational force=mg=77.6 x 9.8=760.48N

So we use this equation:

Fnet = Fn - Fg

ma = Fn - mg

77.6 x 2.1 = Fn – 760.48

162.96N = Fn – 760.48N

923.44 N = Fn

The answer is 923.44 N

6 0

3 years ago

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A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

3 years ago

The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

3 years ago

A particle oscillates in simple harmonic motion, with amplitude A and period T. The particle starts from position x = A. What is

frozen [14]

Answer:

i guess it is D because,

y = A sin(wt)

w = $\frac{2\pi }{T}$ and t = 3T/ 2

now, Y = A sin ( $\frac{2\pi }{T} X \frac{3T}{2}$ )

so, Y = 0radian

Explanation:

8 0

3 years ago