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Nonamiya [84]
3 years ago
7

A vertical block-spring system on earth has a period of 6.0 s. What is the period of this same system on the moon where the acce

leration due to gravity is roughly 1/6 that of earth? A)1.0s B)2.4s C)6.0s D)15s E)36s
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
5 0

Answer:

D) 15s

Explanation:

let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.

the period of a pendulum is given by:

T = 2π√(L/g)

so on earth:

Te = 2π√(L/g1)

     =  6s

on the moon;

Tm = 2π√(L/g2)

since g2 = 1/6 g1 then:

Tm = 2π√(L/(1/6×g1))

      = √(6)×2π√(L/(g1))

and 2π√(L/(g1)) = Te = 6s

Tm = (√(6))×6 = 14.7s ≈ 15s

Therefore, the period of the block-spring system on the moon is 15s.

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Answer:

t1 = t2 + 3.02         V = 41.5

V t1 - 1/2 g t1^2 = V t2 - 1/2 g t2^2

Both stones reach the same height after the specified times

V (t1 - t2) = g/2 (t1^2 - t2^2) = g/2 (t1 - t2) (t1 + t2)

2 V / g = t1 + t2 = 2t1 + 3.02

t1 = V / g - 1.51 = 41.5 / 9.8 -1.51 = 2.72 s

t2 = t1 + 3.02 = 5.74 sec

Check:

41.5 * 2.72 - 4.9 * 2.72^2 = 76.6 m

41.5 * 5.74 - 4.9 * 5.74^2 = 76.8 m

Speed of second stone = 41.5 - 9.8 * 2.72 = 14.8 m/s

7 0
3 years ago
After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc
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Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

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f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

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\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

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d_i = 200 cm

so his far point according to this pair of glass is 200 cm

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3. An astronaut on a space-walk swings a 0.5 kg metal tool in a circular path at the end of a cable 3.6 m long. The astronaut th
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Answer:

Explanation:

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Answer:

d = 11.1 m

Explanation:

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\frac{1}{2} m {v}^{2}  = mgh

Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:

\sin(15)  =  \frac{h}{d}  \\ or \: h = d \sin(15)

Plugging this into the energy conservation equation and cancelling m, we get

{v}^{2}  = 2gd \sin(15)

Solving for d,

d =  \frac{ {v}^{2} }{2g \sin(15) }  =  \frac{ {(7.5 \:  \frac{m}{s}) }^{2} }{2(9.8 \:  \frac{m}{ {s}^{2} })(0.259)}   \\ = 11.1 \: m

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3 years ago
A tennis ball travelling at a speed of 46m/s with a mass of 58kg. Calculate the kinetic<br>energy​
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Answer:

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