Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:
Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
The period of a pendulum is given by
since Karachi is near sea level, g is larger than it is on Mt. Everest. That means the pendulum will have a larger period on Mt. Everest than it would in Karachi.
Answer:
Acceleration of that planet is 30 
.
Given:
initial speed of hammer = 0 
time = 1 s
distance = 15 m
To find:
Acceleration due to gravity = ?
Formula used:
Distance covered by hammer is given by,
s = ut + 
s = distance
u = initial speed of hammer
t = time taken by hammer to reach ground
a = acceleration
Solution:
Distance covered by hammer is given by,
s = ut +
s = distance
u = initial speed of hammer
t = time taken by hammer to reach ground
a = acceleration
u = 0
t = 1 s
s = 15 m
a = g
Thus substituting these value in above equation.
15 = 0 + 
g = 15 × 2
g = 30
Thus, acceleration of that planet is 30 .
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
