Answer:
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ =
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
The work done by a rotating object can be calculated by the formula Work = Torque * angle.
This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.
An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.
The formula is Work = torque * angle.
Torque = 1000 N*m
Angle = [50 revolutions] * [2π radians/revolution] = 100π radians
=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.
Answer:
speed
Explanation:
Speed = distance travelled/time taken
Answer:
L = 1.11 x
m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Explanation:
Solution:
Data Given:
Heat Energy = 52000 J
Dielectric Constant of the plastic Bag = 3.7 = K
Thickness = 2.6 x
m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x
m
Length = ?
So,
we know that,
U = 1/2 C Δ
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ
52000 J = (0.5) x (C) x (
)
C = 0.28 F
And we also know that,
C = 
E = 8.85 x 
K = 3.7
A = 0.20 x L
d = 2.6 x
m
Plugging in the values into the formula, we get:
0.28 = 
Solving for L, we get:
L = 1.11 x
m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.