Answer:
The momentum of block B = 27 Kg m/s
Explanation:
Given,
The initial momentum of block A, MU = 15 Kg m/s
The final momentum of block A, MV = -12 Kg m/s
Consider the block B is initially at rest.
Therefore, the initial momentum of block B, mu = 0
According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.
<em> MU + mu = MV + mv</em>
15 + (0) = (-12) + mv
mv = 15 + 12
= 27 Kg m/s
Hence, the momentum of the block B after impact is, mv = 27 Kg m/s
Answer:
energy carried by the current is given by the pointyng vector
Explanation:
The current is defined by
i = dQ / dt
this is the number of charges per unit area over time.
The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.
But the energy carried by the current is given by the pointyng vector of the electromagnetic wave
S = 1 / μ₀ EX B
It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement
I don’t know what the angle is in your diagram so I used the angle from the vertical.
Answer:
the correct answer is A, the object goes 4 times as far
Explanation:
This is a projectile launching approach. Where the parameter we are controlling is the initial speed and they ask us how far it goes from the initial one. Let's calculate the range with a speed (vo)
R1 = v₀² sin 2θ / g
Now let's double vo, the new speed is
v = 2 v₀
We calculate the scope
R2 = (2v₀)² sin 2θ / g
R2 = 4 v₀² sin 2θ / g
R2 = 4 R1
Therefore the correct answer is A, the object goes 4 times further
Answer:
(a): The frequency of the waves is f= 0.16 Hz
Explanation:
T/4= 1.5 s
T= 6 sec
f= 1/T
f= 0.16 Hz (a)