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aliya0001 [1]
3 years ago
15

As seen from earth how long does it take the moon to rotate on its axis?

Physics
1 answer:
blondinia [14]3 years ago
3 0

It takes approximately 27 days

You might be interested in
A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
Read 2 more answers
1. Why is it easier to change the motion of a large toy car than a small toy car?
insens350 [35]
Because it is acted upon by balanced forces.

Toy cars use a variety of mechanisms to make them go, but they all store up potential energy. Although the elastic material inside is usually steel and not rubber, the principle is the same. By changing the shape of the material (usually a coil of metal) energy is stored and then released as motion.
7 0
3 years ago
What is the centripetal force acting on a 1.5 kg mass moving in a circular path with a centripetal
zloy xaker [14]

The centripetal force acting on a 1.5 kg mass moving in a circular path is 27N

A centripetal force is a net force acting on an object in order to maintain the object's movement in a circular motion.

According  to Newton's first law, it states an object will continue to proceed its movement in a straight line unless acted upon by an external force.

The centripetal force is the external force at work here and  It's the net force that propels the object in a circular motion.

Using the formula for calculating centripetal force:

\mathbf{F_c = mass (m) \times acceleration (a)}

\mathbf{F_c = 1.5 \ kg \times18 \ m/s^2}

\mathbf{F_c = 27 \ N}

Learn more about centripetal force here:

brainly.com/question/11324711?referrer=searchResults

8 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
IgorC [24]

Answer:

The answers to the questions are;

a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.

b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.

c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.

Explanation:

To solve the question, we write out the known variables as follows

1 g of fat = 9.00kcal

Number of steps the student climbs = 95 steps

Height of each step = 0.150 m

Time it takes for the student to reach the top of the stairs = 57.5 s.

Efficiency of human muscles = 20 %

Mass of student, m = 65 kg

a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.

The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..

That is work done, W, =  P. E. = m·g·h

Where:

h = The total height climbed by the student

g = Acceleration due to gravity = 9.81 m/s²

Therefore;

h = Height of each step × Number of steps the student climbs =

  = 0.150 m/(step) × 95 steps = 14.25 m

Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²

                          = 9086.5125 J

We remember that the efficiency of the muscle is 20 %

The formula for efficiency is

Efficiency = \frac{Ene rgy Out put}{Energ y In put} \times 100 %

The work produced by the muscle =  Energy Output = 9086.5125 J

Energy input is given by

\frac{Out put} {Effici ency} = 9086.5125 J/ (0.2) = 45432.5625 J

= 45.432 kJ

From the question, 1 g of fat = 9.00 kcal and

1 kcal = 4186 J

Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J

Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ

To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs \frac{37674 kJ}{45.432 kJ} times to make up the 37674 kJ energy contained in 1 kg of fat

That is  \frac{37674 kJ}{45.432 kJ} =  829.23 times

b. Power is the rate of doing work

That is Power output = \frac{ WorkO utput }{Time} = \frac{9086.5125 J}{57.5 s} = 158.026 watts

c. No as the activity student will have to spend a total time of

829.23 × 57.5 s = 47680.67 s climbing up the stairs alone  and

47680.67 s = ‪13.24 Hours climbing up of which if the time to climb down is the same s climbing up, then we ave total time = 2× ‪13.24 Hours  

= 26.49 hrs = 1.104 days exercising which is not humanly possible.

3 0
3 years ago
3. A difference between copper and plastic is that copper has a much higher
Neko [114]

Answer:

d ............................

6 0
3 years ago
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