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musickatia [10]
3 years ago
14

Imagine you had a bar of gold and decided to cut in half. You repeated this

Physics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

b, an element

Explanation:

this is because you never combined it with any other element in the periodic table and gold is not a compound or mixture.

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I have no clue what to do please help.
Yakvenalex [24]

Answer:

Step one : read the directions

Step two: complete the assignment

Explanation:

3 0
2 years ago
) A steel guitar string with a diameter of 1.00 mm is stretched between supports 80.0 cm apart. The temperature is 0.0°C. (a) Fi
ladessa [460]

Answer: a. Mass per unit length =0.0245kg/m

b. Tension =2.45x10^-8N

C. Tension = 2.45 x10^-8N

Fundamental frequency =200Hz

Explanation:

7 0
3 years ago
if the time is taken to bring a ball to rest from a certain velocity 'v'. is reduced to half, what will be the change in the val
777dan777 [17]

The answer is C) rate of change of momentum. The answer is not initial or final momentum as the start and end points are not changing. On the other hand, the time it takes for the ball to change velocity is. This change relates to the change of momentum. Hope this helped :))

6 0
2 years ago
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
3 years ago
A golf ball comes to rest 1.90 seconds after hitting a net. The force of the net slowed the ball down at a rate of 56.3 m/s^2. W
Nadusha1986 [10]

Answer:

The velocity of the ball when it first hit the net is Vi= 106.97 m/s

Explanation:

Vf= 0

Vi= ?

a= 56.3 m/s²

t= 1.9 s

Vf= Vi - a*t

Vi= a*t

Vi= 106.97 m/s

8 0
2 years ago
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