Answer:
Explanation:
Not enough information.
IF we ASSUME she wants the car to be at LAUNCH LEVEL after 1 second of flight.
THEN
The highest point will have zero vertical velocity and will have taken ½ second to get there. This means that the initial vertical velocity was
v = gt
vy₀ = 9.8(0.5)
vy₀ = 4.9 m/s
vsinθ = vy₀
v = vy₀/sinθ
v = 4.9/sin32
v = 9.2466...
v = 9.2 m/s
Answer:
a
When 
b
When 
![B = [\frac{\mu_o * I }{ 2 \pi R^2} ]* r](https://tex.z-dn.net/?f=B%20%3D%20%20%5B%5Cfrac%7B%5Cmu_o%20%2A%20%20I%20%7D%7B%202%20%5Cpi%20R%5E2%7D%20%5D%2A%20r)
Explanation:
From the question we are told that
The radius is R
The current is I
The distance from the center
Ampere's law is mathematically represented as
![B[2 \pi r] = \mu_o * \frac{I r^2 }{R^2 }](https://tex.z-dn.net/?f=B%5B2%20%5Cpi%20r%5D%20%20%3D%20%20%5Cmu_o%20%20%2A%20%20%5Cfrac%7BI%20r%5E2%20%20%7D%7BR%5E2%20%7D)
When
=>
But when
Force, the unit is Newton, newton is the force to accelerate a mass. so it should be kg m/s^2
joule (J) is equal to Nm not Ns
the unit of work is J and it is correct.
the unit of power is J/s which is equal to W
the unit of of energy is the same with work, which is J which equivalent to kgm2/s2
Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)
