Answer:
Explanation:
Given that
Mass of bowling ball M1=7.2kg
The radius of bowling ball r1=0.11m
Mass of billiard ball M2=0.38kg
The radius of the Billiard ball r2=0.028m
Gravitational constant
G=6.67×10^-11Nm²/kg²
The magnitude of their distance apart is given as
r=r1+r2
r=0.028+0.11
r=0.138m
Then, gravitational force is given as
F=GM1M2/r²
F=6.67×10^-11×7.2×0.38/0.138²
F=9.58×10^-9N
The force of attraction between the two balls is
F=9.58×10^-9N
The broad, slightly dome-shaped volcanoes of Hawaii are sheild volcanoes
Answer:
0.47 J
Explanation:
The elastic potential energy of a spring is given as,
E = 1/2ke²........................ Equation 1
Where E = Elastic potential energy, k = spring constant, e = extension/compression.
Given: k = 15 N/m, e = 0.25 m.
Substitute into equation 1.
E = 1/2(15)(0.25)²
E = 0.46875
E ≈ 0.47 J.
Hence the elastic potential energy stored in the spring = 0.47 J
Answer:
F = 0.64 N
Explanation:
We are given;
Spring constant constant; k = 1.28 N/m
Distance; x = 0.5 m
From Hooke's law, we know that F = kx.
Thus;
F = 1.28 × 0.5
F = 0.64 N
Thus, force it takes to pull the spring back = 0.64 N
Answer: C = Q/4πR
Explanation:
Volume(V) of a sphere = 4πr^3
Charge within a small volume 'dV' is given by:
dq = ρ(r)dV
ρ(r) = C/r^2
Volume(V) of a sphere = 4/3(πr^3)
dV/dr = (4/3)×3πr^2
dV = 4πr^2dr
Therefore,
dq = ρ(r)dV ; dq =ρ(r)4πr^2dr
dq = C/r^2[4πr^2dr]
dq = 4Cπdr
FOR TOTAL CHANGE 'Q', we integrate dq
∫dq = ∫4Cπdr at r = R and r = 0
∫4Cπdr = 4Cπr
Q = 4Cπ(R - 0)
Q = 4CπR - 0
Q = 4CπR
C = Q/4πR
The value of C in terms of Q and R is [Q/4πR]
