Answer:
The angle between the electric field and the axis of the filter is 54⁰
Explanation:
Apply the equation for intensity of light through a polarizer.
where;
I is the intensity of the transmitted light
I₀ is the intensity of the incident light
θ is the incident angle
If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;
Therefore, the angle between the electric field and the axis of the filter is 54⁰
As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
Answer:
The tension in the rope is 262.88 N
Explanation:
Given:
Weight 
N
Length of rope 
m
Initial speed of ball 
For finding the tension in the rope,
First find the mass of rod,
( 
)
kg
Tension in the rope is,
N
Therefore, the tension in the rope is 262.88 N
<span>1) The differential equation that models the RC circuit is :
(d/dt)V_capacitor </span>+ (V_capacitor/RC) = (V_source/<span>RC)</span>
<span>Where the time constant of the circuit is defined by the product of R*C
Time constant = T = R*C = (</span>30.5 ohms) * (89.9-mf) = 2.742 s
2) C<span>harge of the capacitor 1.57 time constants
1.57*(2.742) = 4.3048 s
The solution of the differential equation is
</span>V_capac (t) = (V_capac(0) - V_capac(∞<span>))e ^(-t /T) + </span>V_capac(∞)
Since the capacitor is initially uncharged V_capac(0) = 0
And the maximun Voltage the capacitor will have in this configuration is the voltage of the battery V_capac(∞) = 9V
This means,
V_capac (t) = (-9V)e ^(-t /T) + 9V
The charge in a capacitor is defined as Q = C*V
Where C is the capacitance and V is the Voltage across
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /T) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V
V_capac (4.3048 s) = (-9V)e ^(-4.3048 s /2.742 s) + 9V = -1.87V +9V
V_capac (4.3048 s) = 7.1275 V
Q (4.3048 s) = 89.9mF*(7.1275V) = 0.6407 C
3) The charge after a very long time refers to the maximum charge the capacitor will hold in this circuit. This occurs when the voltage accross its terminals is equal to the voltage of the battery = 9V
Q (∞) = 89.9mF*(9V) = 0.8091 C
Average speed = total distance / total time.
That's 172 km / 5 hr = 34.4 km//hr
