There is still air inside of a house, which is pushing the roof upwards, so the forces are equal and the roof is not crushed.
Answer:
The radius of the gold nucleus is 7.1x10⁻¹⁵m
Explanation:
The nearest distance is:
(eq. 1)
Where
z = atomic number of gold = 79
e = electron charge = 1.6x10⁻¹⁹C
k = electrostatic constant = 9x10⁹Nm²C²
energy of the particle = 32 MeV = 5.12x10⁻¹²J
At the potential energy is zero, all the energy will be kinetic energy:

Where
m = 4 mp = mass of proton

Replacing in equation 1

Answer:
d. equal to one-fourth the acceleration at the surface of the asteroid.
Explanation:
The explanation is attached as a picture with this answer
Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.
as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.
As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.
It is taken 2r for the top is the path. hence we obtain (1/4)g as our answer.
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )