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MatroZZZ [7]
2 years ago
8

Suppose Earth orbited a star whose mass was double the mass of the sun. If the radius of Earth’s orbit remained the same as it i

s now, then compared with the gravitational force between Earth and the sun, the gravitational force between Earth and the star would be—
half as much
the same
two times as much
four times as much
Physics
1 answer:
Vaselesa [24]2 years ago
7 0

Answer:

Two times as much

Explanation:

The equation for gravitational force is: Fg = GMm/r^2 with G being the universal gravitational constant.

So to make things easier we'll set r equal to 1 since it's a constant as well as G.

Then we're left with Fg=Mm with M being the mass of the sun and m being the mass of the earth.

So if m is constant and supposedly equals 1 then Fg=M so Fg is proportional to M therefore if M doubles then Fg doubles.

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Rainbow [258]
Earth-like planet in another solar system
Hope this helps!!
8 0
3 years ago
A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa
kotykmax [81]
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is v_i=180~m/s, the final velocity is v_i=0~m/s, and the total time of the motion is \Delta t=0.02~s, so the acceleration is given by
a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2 
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m
So, the bullet penetrates the sandbag 1.8 meters.
5 0
3 years ago
Acceleration is defined as the rate of change for which of the following?
Snowcat [4.5K]
<span>B. velocity .................</span>
6 0
3 years ago
Read 2 more answers
30 points!
vlada-n [284]

Answer:

Since strong nuclear forces involve only nuclear particles (not electrons, bonds, etc)  items 3 and 4 are eliminated.

Again item 2 refers to bonds between atoms and is eliminated.

This leaves only item 1.

Nuclear forces are very short range forces between components of the nucleus.

Weak nuclear forces are trillions of times smaller than strong forces.

Gravitational forces are much much smaller than the weak nuclear force.

6 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
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