Answer:
Explanation:
Impulse of a force is measured by force x time or F X t
Impulse also equals change in momentum or
F x t = m v₂ - m v₁
The given case is as follows
in the first case
F x t = mv - o = mv
F = mv / t
in the second case
F₁ x 4 t = mv
F₁ = 1/4 x mv /t
F₁ = F / 4
option a) is correct .
iii )
In the last case
F₂ X t = m v/2 -0
F₂ = 1/2 x mv / t
= 1/2 x F
F₂ = F/2
Option e ) is correct.
The answer is: 120V
Power is the rate at which energy is supplied/transformed in time:
we can write:
V ddp in Volts represents Energy/Charge i.e. energy carried by each coulomb;
I current in Amperes represents Charge/time or coulombs passing each seconds.
combining them we have:
Power = energy/time = V • 1
or
1200 = V ⋅ 10
V = 1200/10 = 120V
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
Explanation:
It is given that,
Magnitude of charge,
It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.
Magnetic field,
Velocity,
We need to find the magnitude of force on the charge. Magnetic force is given by :
<em>Since</em>,
So, the force acting on the charge is and is moving in positive z axis. Hence, this is the required solution.