For Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
Rewriting the acceleration as the increment of velocity
in a time
:
, F becomes
But given the definition of momentum:
, then
represents the momentum change. So we can rewrite F as
And re-arranging the formula we can calculate the value of the change in momentum:
<span>When a person lifts the block, the block has more potential energy. Therefore the person does positive work on the block.
work = m g h
work = (4.5 kg) (9.80 m/s^2) (1.2 m)
work = 52.92 joules
The person's work on the block is 52.92 joules
When the block is being raised, the force of gravity opposes the motion. Therefore the force of gravity does negative work on the block.
work = - (force) (h)
work = - m g h
work = -(4.5 kg) (9.80 m/s^2) (1.2 m)
work = -52.92 joules
The work done by the force of gravity on the block is -52.92 joules
Note that when the block is moved horizontally, the potential energy does not change. Therefore there is no work done on the block when it moves horizontally (we are assuming that the kinetic energy does not change).</span>
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
= 1250W
d) Final peak=
P= Ik/e
= 
P = 2.5 × 10⁷W
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration
. Calling h its height at t=0, the height at time t is given by
We are told thatn when
the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
2) The speed of the grapefruit at time t is given by
where
is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
