Answer: It would increase.
Explanation:
The equation for determining the force of the gravitational pull between any two objects is:

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.
Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.
Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.
Answer:
The total kinetic energy of both particles is 
Explanation:
Given that,
Kinetic energy of nucleus
Kinetic energy of proton 
Radius of proton 
We need to calculate the final potential energy
Using formula of final potential energy

Put the value into the formula


We need to calculate the initial energy of both the particles
Using formula of energy



We need to calculate the total kinetic energy of both particles
Using conservation of energy





Hence, The total kinetic energy of both particles is 
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

Hi, thank you for posting your question here at Brainly.
To compute for the change in potential energy, the equation would be:
delta PE = mg*delta h
delta PE = 0.5*9.81*(2-1.8)
delta Pe = 0.98 J
The potential energy is converted to kinetic energy.
Answer:
Angular displacement of the turbine is 234.62 radian
Explanation:
initial angular speed of the turbine is



similarly final angular speed is given as



angular acceleration of the turbine is given as

now we have to find the angular displacement is given as


