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jeka57 [31]
4 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]4 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 plece of titanium has a mass of 67.5g and a volume of 15cm<br> What is the density?
sveta [45]

Answer:

4.5g/cm^3

Explanation:

Here, Mass(m)=67.5g

         Volume(v)=15cm^3

Now, According to formula,

        Density(p)=m/v

                         =67.5/15

                         =4.5g/cm^3

8 0
3 years ago
Deanna stirred a teaspoon of sugar into a glass of warm water. The sugar completely dissolved in the water. Select 3 statements
larisa [96]

Answer:

B, C, F

Explanation:

B: Sugar can be separated from the water by evaporating the water. This will leave large chunks of sugar.

C: Sugar gets spread out among the water.

F: Sugar water is a homogeneous <u>mixture. </u>Can't see the individual components because of the dissolving.

Hoped this helped! :)

8 0
3 years ago
A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete
lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ F_x=mgSin \theta=ma

⇒ a_x=gSin \theta

       =9.8 \ Sin35^{\circ}

       =5.62 \ m/s^2

(b)

If friction present,

⇒ F_x=mgCos \theta

⇒ F_x=mgSin \theta-\mu_r mgCos \theta

⇒ a=gSin \theta-\mu_rgCos \theta

      =9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}

      =5.46 \ m/s^2

8 0
3 years ago
Can one of yall help???
Art [367]
1,200 x 3.0 = 3600N
8 0
3 years ago
Read 2 more answers
How do we use light in space?
bulgar [2K]

Answer:

They use LED lights.

Explanation:

Hope this helps

-A Helping Friend (mark brainliest pls)

5 0
3 years ago
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