Answer:5101.35v
Explanation:
Radius of gold nucleus=7.3×10-15m and a charge of +79e
Q= 79e
e=1.6×10^-19
q= +2e
The nucleus is considered as the point charge where the potential energy between the charges are
U = 1/(4×3.142×Eo)×(qQ)/r
Where r is distance between the charges and the nucleus
r=R+d
V=U/q
U= 1/(4×3.142×Eo)×Q/r
V= 1/(4×3.142×Eo)×Q/(r+d)
9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)
V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)
V= 9×10^9×(5.67×10^-14)= 5101.35v
The net force applied to the object equals the mass of the object multiplied by the amount of its acceleration." The net force acting on the soccer ball is equal to the mass of the soccer ball multiplied by its change in velocity each second (its acceleration).
Work of the force = 10 N
Time required for the work = 50 sec
Height = 7 m
We are given with the value of work and time in the question.
Substitute the values in the formula of power and then you'll get the power required.
We know that,
w = Work
p = Power
t = Time
By the formula,
Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,
p = 70/50
p = 1.4 watts
Therefore, the power required is 1.4 watts.
Hope it helps!
This assumes that the wave has velocity c (is light).
Answer:
The height of the image is, h' = 6.0 cm
The image is erect.
Explanation:
Given data,
The object distance, u = -5 cm
The focal length of convex lens, f = 10 cm
The object height, h = 3 cm
The lens formula,



v = -10 cm
The magnification factor of lens

m = 2



h' = 6 cm
The height of the image is, h' = 6 cm
The image is erect.