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3 years ago

If earth began to shrink, but its mass remained the same, what would happen to the value of g on earth's surface?

1 answer:

OverLord2011 [107]3 years ago

3 0

<span>The estimation of g on the surface of the earth relies upon mass and range, truth be told, g = GM/R^2 where G is the newton's gravitational constants, M is the mass of the earth and R the range . Should there be any occassion that would cause R to be diminished, the estimation of g would increment as the opposite square of R on the fact that R is divided, g would be 4 times the original value</span>

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A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?

Mashcka [7]

Answer:

<em>v=40 m/s south</em>

Explanation:

<u>Momentum </u>

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

$p=m.v$

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

$\displaystyle v=\frac{p}{m}$

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

$\displaystyle v=\frac{6}{0.15}=40\ m/s$

The velocity is directed to the south

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3 years ago

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2 years ago

15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m

Rudik [331]

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters

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3 years ago

Read 2 more answers

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T

mart [117]

<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;

Charge of electron e= 1.602 × 10^-19 C;

kinetic energy E= 2.7 MeV

= 2.7 × 10^6 × 1.602 × 10^-19 J;

= 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E = 4.32 × 10^-13 Joules

r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

= 2.775 × 10^7 rad /sec

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3 years ago