Newton's first law can be taken to mean that if something is moving it tends to keep moving. if at rest it tends to stay at rest.
so, in a car, you and the car are both moving, say at constant speed. Now you're not actually connected to the car as in clamped to it, not yet at least. You're simply sitting in it at rest with respect to it.
but, someone slams on the brakes for whatever reason. The car slows down/stops. what do you do ? well, you would keep going. and moving a few feet in a car can be dangerous, esp if you're moving at high speed. Unless of course you're clamped to the seat, and the seat is clamped to the car and the car is clamped together. then when the car brakes, yes you'll feel the braking effect, but the belt will restrict your movement, keeping you safe, if shocked and bruised.
Answer:
systems
Explanation:
they are a bunch of things functioning together
Answer:
F = 4.147 × 10^23
v = 1.31 × 10^4
Explanation:
Given the following :
mass of Jupiter (m1) = 1.9 × 10^27
Mass of sun (m2) = 1.99 × 10^30
Distance between sun and jupiter (r) = 7.8 × 10^11m
Gravitational force (F) :
(Gm1m2) / r^2
Where ; G = 6.673×10^-11 ( Gravitational constant)
F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2
F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)
F = (25.231 × 10^46) / (60.84 × 10^22)
F = 3.235 × 10^(46 - 22)
F = 0.4147 × 10^24
F = 4.147 × 10^23
Speed of Jupiter (v) :
v = √(Fr) / m1
v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)
v = √32.3466 × 10^(23+11) / 1.9 × 10^27
v = √32.3466× 10^34 / 1.9 × 10^27
v = √17. 023 × 10^34-27
v = √17.023 × 10^7
v = 13047.221
v = 1.31 × 10^4
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
