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2 years ago

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How was the gravitational constant G first determined

1 answer:

Oliga [24]2 years ago

6 0

<span>In equation form, this is often expressed as follows: The constant of proportionality in this equation is G - the universal gravitation constant. The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance.</span>

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How does the mass of an object affect its acceleration during free fall

ira [324]

The mass of an object has no effect whatsoever on the object's
acceleration during free-fall. If there is no air resistance to interfere
with the natural effects of gravity, then a feather and a battleship ...
dropped at the same time ... fall together, and hit the ground at the
same time.

8 0

3 years ago

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You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil

Step2247 [10]

Answer:

C. it will not change.

Explanation:

While combing, the rubbing of the comb with the hair, transfer of electron takes place from the hair to the comb and the comb becomes negatively charged. But, this transfer of electron does not make any considerable change in the mass of the hair. This is because the mass of an electron is highly negligible. Now, neglecting the mass of an electron, the transfer of the electrons from the hair to the comb makes charging of the comb, but no loss of mass in the hair. So, the mass of hair will no change.

7 0

3 years ago

A 88.6-kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.9kg . (Use 9.80m/s2 for the gravitational acce

Dafna11 [192]

Answer:

Tension maximum =1131.9 N

Tension minimum =868.28 N

Tension at 3/4= 1065.995 N

Explanation:

a)

Given Mass of wrecking ball M1=88.6 Kg

Mass of the chain M2=26.9 Kg

Maximum Tension Tension max=(M1+M2) × (9.8 m/s²)

=(88.6+26.9) × (9.8 m/s²)

=115.5 × 9.8 m/s²

Tension maximum =1131.9 N

b)

Minimum Tension Tension minimum=Mass of the wrecking ball only × 9.8 m/s²

=88.6 × 9.8 m/s²

Tension minimum =868.28 N

c)

Tension at 3/4 from the bottom of the chain =In this part you have to use 75% of the chain so you have to take 3/4 of 26.9

= (3/4 × 26.9)+88.9) × 9.8 m/s²

= (20.175+88.6) × 9.8 m/s²

=(108.775) × 9.8 m/s²

=1065.995 N

6 0

3 years ago

A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What

kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let $p_{1}$ = momentum of the 1st ball

$p_{2}$ = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

$(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}$

or

$(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}$

Since we are dealing with identical balls, all the m terms cancel out so we are left with

$(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}$

Putting in the numbers, we get

$1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}$

$= > (v_{2})_{f} \cos \theta_{2} = 0.703$

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

$0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}$

or

$= > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412$

Taking the ratio of the sine equation to the cosine equation, we get

$\frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586$

or

$\theta_{2} = { \tan}^{ - 1} (0.586) = 30.4$

Solving now for $(v_{2})_{f}$ ,

$(v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s}$

3 0

2 years ago

What is equall to E=MC

NeTakaya

Answer: Nothing is, if im not mistake

Explanation:

6 0

2 years ago

Read 2 more answers