Answer: a) the force will be repulsive
b) the ratio of the new force to the old force will be 2
c) O
Explanation:
a) since charge -Q is moved from A to B, this implies that sphere A is negatively charged. The two spheres are now negatively charged and will repel themselves.
b) initial force will be -q(-Q)/d2
Adding extra charge -Q will cause change on B to become -2Q
The new force will be - 2Q(-q)/d2
Dividing new force by old force will give 2
C) if B is neutralized, the net charge becomes 0 and there will be no force on it.
Answer:
24.531 m
Explanation:
t = Time taken = 1.7 s
u = Initial velocity = 6.1 m/s
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
The initial height of the rock above the ground is 24.531 m
Answer:
h = 13.06 m
Explanation:
Given:
- Specific gravity of gasoline S.G = 0.739
- Density of water p_w = 997 kg/m^3
- The atmosphere pressure P_o = 101.325 KPa
- The change in height of the liquid is h m
Find:
How high would the level be in a gasoline barometer at normal atmospheric pressure?
Solution:
- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.
- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:
P = S.G*p_w*g*h
Where, h = P / S.G*p_w*g
- Input the values given:
h = 101.325 KPa / 0.739*9.81*997
h = 13.06 m
- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.
Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
