maximum static friction acting on the object will be

plug in all values

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force
So here it is given that applied force is 20 N
so here object will not move due to this force and it will remain at rest always
due to this applied force
Answer:
Earth attract the Moon with a force that is greater.
Explanation:
According to the law of gravitation, the gravitational force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematically, F1 = Gm1m2/r²... 1
Let m1 be the mass of the earth and m2 be that of the moon
If the Earth is much more massive than is the Moon, the new force of attraction between them will become;
F2= G(2m1)m2/r²
F2 = 2Gm1m2/r² ... (2)
Dividing eqn 1 by 2 we have;
F1/F2 = (Gm1m2/r²)÷(2Gm1m2/r²)
F1/F2 = Gm1m2/r²×r²/2Gm1m2
F1/F2 = 1/2
F2=2F1
This shows that that the earth will attract the moon by a force 2times the initial force of the masses(i.e a much greater force)
Answer:
0.2 m
Explanation:
PHASE 1
First, we calculate the distance the tongue moved in the first 20 ms (0.02 secs). We use one of Newton's equations of linear motion:

where u = initial velocity = 0 m/s
a = acceleration = 
t = time = 0.02 s
Therefore:

PHASE 2
Then, for the next 30 ms (0.03 secs), we use the formula:

This speed is the same as the final velocity of the tongue after the first 20 ms.
This can be obtained by using the formula:

Therefore:
distance = 5 * 0.03 = 0.15 m
Therefore, the total distance moved by the tongue in the 50 ms interval is:
0.05 + 0.15 = 0.2 m
Answer:
0.0109 m ≈ 10.9 mm
Explanation:
proton speed = 1 * 10^6 m/s
radius in which the proton moves = 20 m
<u>determine the radius of the circle in which an electron would move </u>
we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)
For proton :
Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )
For electron:
re = Me*V/ qE * B -------- ( 2 )
Next: take the ratio of equations 1 and 2
re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )
∴ re ( radius of the electron orbit )
= ( Me / Mp ) rp
= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20
= ( 5.45 * 10^-4 ) * 20
= 0.0109 m ≈ 10.9 mm