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Vinil7 [7]
3 years ago
13

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

orce on the sun._______________The earth exerts some force on the sun, but less than 3.53×1022N because the earth, which is exerting the force, is so much less massive than the sun._______________The earth exerts 3.53×1022N of force on the sun, exactly the same amount of force the sun exerts on the earth found in Part A._______________The earth exerts more than 3.53×1022N of force on the sun because the sun, which is experiencing the force, is so much more massive than the earth._______________
Physics
1 answer:
Sholpan [36]3 years ago
4 0

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.  

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

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Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the
stealth61 [152]

Answer:

\frac{50}{\pi }Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

<u><em>Now,</em></u>

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

<em><u>By comparing equations (i) and (ii) the following holds;</u></em>

V = 230

ω = 100

Ф = 0

<em><u>But;</u></em>

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = \frac{50}{\pi }Hz

Therefore, the frequency of the voltage is \frac{50}{\pi }Hz

7 0
3 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

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