To do this you want to solve for one variable at a time. So we want to cancel out a variable. Lets cancel x. I will multiply the first equation by the number 4 to get 4y=4x-16.
Now lets solve equation 2 for y, giving
-3y=-4x+3 now add equation 1 to equation 2
Y =-13
Now plug that back in to either
-13=x-4
X=-9
So the answer is (-9,-13)
Answer:
Capacitive reactance of the capacitor is 68 ohms
Explanation:
It is given that,
Capacitance, 
Frequency, 
Capacitive reactance is given by :
or
So, the capacitive reactance of the capacitor is 68 ohms. Hence, this is the required solution.
Answer:
Explanation:
a) R = V/I = 0.7/0.022 = 32Ω
b) R = V/I = 0.6/0.005 = 120Ω
c) No...Current change is not linear with voltage change.
Complete Question
A truck going 15 km/h has a head-on collision with a small car going 30 km/h. Which statement best describes the situation?
A. the truck has the greater change of momentum because it has the greater mass
B. the car has the greater change of momentum because it has the greater speed
C. neither the car nor the truck changes its momentum in the collision because momentum is conserved
D. they both have the same change in magnitude of momentum because momentum is conserved
E. none of the above is necessarily true
Answer:
D. They both have the same change in magnitude of momentum because momentum is conserved
Explanation:
In order to get a good understanding of the solution above we define some
concetps
Momentum
This is defines quantified motion and can be mathematically represented as
Momentum = Mass of the body × Velocity of the body
According to the Law of conservation of momentum states that when two particles collide together in a system that is being isolated that their total momentum before and after their collision is equal this means that the momentum lost by the truck would be the same as the momentum gained by the small car
Answer:
B = 1.1413 10⁻² T
Explanation:
We use energy concepts to calculate the proton velocity
starting point. When entering the electric field
Em₀ = U = q V
final point. Right out of the electric field
em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
q V = ½ m v²
v = 
we calculate
v = 
v = 
v = 25.15 10⁴ m / s
now enters the region with magnetic field, so it is subjected to a magnetic force
F = m a
the force is
F = q v x B
as the velocity is perpendicular to the magnetic field
F = q v B
acceleration is centripetal
a = v² / r
we substitute
qvB =1/2 m v² / r
B = v
we calculate
B = 
B = 1.1413 10⁻² T
