Answer:
B
Explanation:
Given:-
- The charge of the test particle q = 3.0 * 10^-9 C
- The force exerted by the metal sphere F = 6.0 * 10^-5 N
Find:-
The magnitude and direction of the electric field
strength at this location?
Solution:-
- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:
F = E*q
- Using the data given we can determine E:
E = F / q
E = (6.0 * 10^-5) / (3.0 * 10^-9)
E = 20,000 N/C
- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.
When vapor pressure equals atmospheric pressure, it's called boiling point of that liquid.
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
