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3 years ago

Which sentence in the passage describes retrograde motion of a planet

Physics

1 answer:

Soloha48 [4]3 years ago

4 0

Answer:

what is the passage? I don’t see it

Explanation:

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How much work did the movers do (horizontally) pushing a 46.0-kgkg crate 10.5 mm across a rough floor without acceleration, if t

-BARSIC- [3]

Answer:

7.1 J

Explanation:

From the question,

Work done by the mover = work done in pushing the crate + work done against friction

W = W'+Wf................. Equation 1

W = mgd+mgμd............ Equation 2

W = mgd(1+μ)................ Equation 3

Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of friction.

Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5

constant: g = 9.8 m/s²

Substitute these values into equation 3

W = 46×9.8×0.0105(1+0.5)

W = 7.1 J

7 0

3 years ago

How do the tension of the cord and the force of gravity affect a pendulum

azamat

The time period of the pendulum is affected by the acceleration due to gravity. The tension does not have any effect on the time period of the pendulum and also the mass of the bob does not effect the time period of the pendulum.

$T = 2 \pi \sqrt{\frac{l}{g}}$

Hence, if the gravity increases then the time period of the pendulum will decreases and it will swing faster.

5 0

3 years ago

What is the meaning of physic

irina [24]

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

5 0

2 years ago

Modern scientists say that electrons are found in a ...

bulgar [2K]

I think it might be a or d i hope i have helped :)

8 0

3 years ago

Read 2 more answers

A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a

Nimfa-mama [501]

Answer:

$F_0 = 393 N$

Explanation:

As we know that amplitude of forced oscillation is given as

$A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}$

here we know that natural frequency of the oscillation is given as

$\omega_0 = \sqrt{\frac{k}{m}}$

here mass of the object is given as

$m = \frac{W}{g}$

$\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}$

$\omega_0 = 8.48 rad/s$

angular frequency of applied force is given as

$\omega = 2\pi f$

$\omega = 2\pi(10.5) = 65.97 rad/s$

now we have

$0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}$

$F_0 = 393 N$

6 0

3 years ago