An equal and opposite force acting upwards
Answer:
Right shoe
Explanation:
Let the mass and velocity of incoming puck be m and v respectively.
Momentum of the colliding puck will be mv
In case of first case , the momentum of puck becomes zero so change in momentum after collision with left shoe
= mv - 0 = mv
If time duration of collision be t
rate of change of momentum
= mv / t
This is the force exerted by puck on the left shoe .
Now let us consider collision with right shoe
momentum after collision with right shoe
- mv
change in momentum
= mv - ( - mv ) = 2mv
If time duration of collision be t
rate of change of momentum
= 2mv / t
This is the force exerted by puck on the right shoe .
Since the force on the right shoe is more , this shoe will have greater speed
after collision.
Answer:
Explanation:
Given an RL circuit
A voltage source of.
V = 108V
A resistor of resistance
R = 1.1-kΩ = 1100 Ω
And inductor of inductance
L = 34 H
After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor
A. Time the inductor current will reduce to 12% of it's initial current
Let the initial charge current be Io
Then, final current is
I = 12% of Io
I = 0.12Io
I / Io = 0.12
The current in an inductor RL circuit is given as
I = Io ( 1—exp(-t/τ)
Where τ is time constant and it is given as
τ = L/R = 34/1100 = 0.03091A
So,
I = Io ( 1—exp(-t/τ))
I / Io = ( 1—exp(-t/τ))
Where I/Io = 0.12
0.12 = 1—exp(-t/τ)
0.12 — 1 = —exp(-t/τ)
-0.88 = -exp(-t/0.03091)
0.88 = exp(-t/0.03091)
Take In of both sides
In(0.88) = In(exp(-t/0.03091)
-0.12783 = -t/0.030901
t = -0.12783 × 0.030901
t = 3.95 × 10^-3 seconds
t = 3.95 ms
B. Energy stored in inductor is given as
U = ½Li²
So, the current at this time t = 3.95ms
I = Io ( 1—exp(-t/τ))
Where Io = V/R
Io = 108/1100 = 0.0982 A
Now,
I = Io ( 1—exp(-t/τ))
I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))
I = 0.0982(1—exp(-0.12783)
I = 0.0982 × 0.12
I = 0.01178
I = 11.78mA
Therefore,
U = ½Li²
U = ½ × 34 × 0.01178²
U = 2.36 × 10^-3 J
U = 2.36 mJ
Answer:
The velocity of the boat after the batman lands in it is +9.26 m/s
Explanation:
Applying the law of conservation of momentum,
Total momentum before collision = Total momentum after collision.
Note: The collision between the Batman and the boat is an inelastic collision.
m'u'+mu = V(m+m').................... Equation 1
Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.
make V the subject of equation 1
V = (m'u'+mu)/(m+m')............... Equation 2
Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.
Substitute these values into equation 2
V = [(96.1×0)+(458×11.2)]/(96.1+458)
V = 5129.6/554.1
V = +9.26 m/s
