Answer:
a block sliding down a ramp,a leaf blowing across a field
You need to observe the car at two different times.
-- The first time:
You write down the car's speed, and the direction it's pointing.
-- The second time:
You write down the car's speed and the direction it's pointing, again.
You take the data back to your lab to analyze it.
-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.
-- You compare the first and second direction. If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.
(Remember, "acceleration" doesn't mean "speeding up".
It means any change in speed or direction of motion.)
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
There is no equation here
Answer:
Explanation:
First of all we shall find the velocity at equilibrium point of mass 1.2 kg .
It will be ω A , where ω is angular frequency and A is amplitude .
ω = √ ( k / m )
= √ (170 / 1.2 )
= 11.90 rad /s
amplitude A = .045 m
velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s
= .5355 m /s
At middle point , no force acts so we can apply law of conservation of momentum
m₁ v₁ = ( m₁ + m₂ ) v
1.2 x .5355 = ( 1.2 + .48 ) x v
v = .3825 m /s
= 38.25 cm /s
Let new amplitude be A₁ .
1/2 m v² = 1/2 k A₁²
( 1.2 + .48 ) x v² = 170 x A₁²
( 1.2 + .48 ) x .3825² = 170 x A₁²
A₁ = .0379 m
New amplitude is .0379 m
