Answer:
So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.
Explanation:
A option.
Answer:
Time is 14.8 s and cannot landing
Explanation:
This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing are the same
Calculate the time to stop, where it has zero speed
Vf² = Vo² + a t
t = - Vo² / a
t = - 110²/(-7.42)
t = 14.8 s
This is the time it takes to stop the jet
Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero
Vf² = Vo² + 2 to X
X = -Vo² / 2 a
X = -110² / 2 (-7.42)
X = 815.4 m
Since this distance is greater than the length of the runway, the jet cannot stop
Let's calculate the speed you should have to stop on a track of this size
Vo² = 2 a X
Vo = √ (2 7.42 800)
Vo = 109 m / s
It is conclusion the jet must lose some speed to land on this track
Answer:
7.7 km 26°
Explanation:
The total x component is:
x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87
The total y component is:
y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38
The magnitude is:
d = √(x² + y²)
d = 7.7 km
The direction is:
θ = atan(y/x)
θ = 26°
Answer: 45.3°
Explanation:
Given,
Length of ladder = l
Weight of ladder = w
Coefficient of friction = μs = 0.495
Smallest angle the ladder makes = θ
If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,
N1 = mg and
N2 = μmg
Moment at a point A in the clockwise direction is
N2 Lsinθ - mg.(L/2).cosθ = 0
μmgLsinθ - mg.(L/2).cosθ = 0
μmgLsinθ = mg.(L/2).cosθ
μsinθ = cosθ/2
sin θ / cos θ = 1 / 2μ
Tan θ = 1 / 2μ
Substituting the value of μ = 0.495, we have
Tan θ = 1 / 2 * 0.495
Tan θ = 1 / 0.99
Tan θ = 1.01
θ = tan^-1(1.01)
θ = 45.3°
The answer is to u question is: C
