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VladimirAG [237]
3 years ago
11

Calculate the acceleration due to gravity on venus. the radius of venus is about 6.06 x 106 m and its mass is 4.88 x 1024 kg.

Physics
1 answer:
Elena-2011 [213]3 years ago
7 0
The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of venus
r is the radius of venus

Substituting the values,

a = (6.6726 x 10⁻¹¹ N-m²/kg²)(4.88x10²⁴ kg)/(6.06x10⁶ m)²
<em>a = 8.87 m/s²</em>
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A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.
MaRussiya [10]

Answer:

0.76 s

Explanation:

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Length of rod,L=20 cm=\frac{20}{100}=0.20m

1 m=100 cm

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Mass of ball,m=19 g=\frac{19}{1000}=0.019 kg

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a  pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

I_{rod-clay}=I_{rod}+I_{clay}

I_{rod-clay}=\frac{1}{3}ML^2+mL^2

Substitute the values then we get

I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2

I_{rod-clay}=3.29\times 10^{-3} kgm^2

Now, the center of mass of the combination of the rod and clay is given by

y=\frac{Md_1+md_2}{M+m}

Substitute d_1=\frac{L}{2}=Distance between pivot and the center of the rod

d_2=L=The distance  between rod and clay

Using the formula

y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}

y=0.1091 m

Time period of the oscillation of the system of the rod and the clay is given by

T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}

g=9.8m/s^2

Using the formula

Time period=2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}

Time-period=0.76 s

Hence, the period =0.76 s

8 0
3 years ago
Suppose a piano tuner hears 2 beats per second when listening to the combined sound from her tuning fork and the piano note bein
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Answer:

further tightening is required.

Explanation:

The beat created / sec = difference of frequencies

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after tightening beat heard  = 1

difference  of frequencies decreases because frequency of tuning fork was higher than piano sound.

on further tightening  difference decreases because tightening increases the frequency  of piano hence   further  tightening is required for resonance.

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