Answer:
The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.
Explanation:
We will use the equations of motion for this.
u = initial velocity of the rock = 22 m/s
g = acceleration due to gravity = -9.8 m/s²
y = vertical position of the rock at a time t = 9 m
y₀ = initial height of the rock = 25 m
t = time it takes for the rock to reach height of 9 m.
(y-y₀) = ut + 0.5gt²
(9 - 25) = 22t + 0.5(-9.8)t²
- 14 = 22t - 4.9t²
4.9t² - 22t - 14 = 0
solving this quadratic equation,
t = 5.055 s or - 0.565 s
Since time cannot be negative,
t = 5.055 s = 5.06 s
Hope this Helps!!!
Electrical power, in watts = (voltage, in volts) x (current, in Amperes)
Complete Question
The complete question is shown on the first uploaded image (reference for Photobucket )
Answer:
The electric field is
Explanation:
From the question we are told that
The linear charge density on the inner conductor is
The linear charge density on the outer conductor is
The position of interest is r = 37.3 mm =0.0373 m
Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )
Generally according to Gauss Law
=>
substituting values
The negative sign tell us that the direction of the electric field is radially inwards
=>