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jeka57 [31]
3 years ago
15

You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.

Will the period be the same when the buggy is used on the moon? Explain.
Physics
1 answer:
Blizzard [7]3 years ago
3 0

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

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A crane lifts an air conditioner to the top of a building. If the building is 12 m high, and the air conditioner has a mass of 2
Pepsi [2]

Work needed = 23,520 J

<h3> Further explanation </h3>

Given

height = 12 m

mass = 200 kg

Required

work needed by the crane

Solution

Work is the transfer of energy caused by the force acting on a moving object  

Work is the product of force with the displacement of objects.  

Can be formulated  

W = F x d  

W = Work, J, Nm  

F = Force, N  

d = distance, m  

F = m x g

Input the value :

W = mgd

W = 200 kg x 9.8 m/s²x12 m

W = 23520 J

8 0
2 years ago
Why is it advantageous to have two Eyes?​
Andru [333]

Answer:para enxergar

Explanation:

6 0
2 years ago
Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of
Anastasy [175]

Answer:

<em>d. 268 s</em>

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

\displaystyle v=\frac{d}{t}

Where  

v = Speed of the object

d = Distance traveled

t  = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

\displaystyle t=\frac{d}{v}

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

\displaystyle t=\frac{5}{67}

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0
3 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × \sqrt{\frac{m}{k} }   ................2

put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

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