Answer:
The capacite is C=5.32 uF using the equations of voltage and energy in capacitance
Explanation:
The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J
Using:
Voltage in this case is the energy dissipated so
Using the equation to find capacitance
F
C= 5.32 uF because u is the symbol for micro that is equal to 
Answer:
Stage one of sleep, also known as the transitional phase, occurs when one finds themselves floating in and out of consciousness. During this NREM stage, you may be partially awake while your mind begins to drift off. This period of drowsiness eventually leads to a light sleep
Explanation:
i found it on google
v = initial velocity of launch of the stone = 12 m/s
θ = angle of the velocity from the horizontal = 30
Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.
v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s
a = acceleration of the stone = - 9.8 m/s²
t = time of travel = 4.8 s
Y = vertical displacement of stone = vertical height of the cliff = ?
using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²
Y = - 84.1 m
hence the height of the cliff comes out to be 84.1 m
Answer:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² = 
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo = ![v_{oy}² /2 g y- 0 = 10.0²/2 9.8 y - 0 = 5.10 m The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system y₂ = 5.1 + 44 y₂ = 49.1 m Let's use the other equation to find the time [tex]v_{y}](https://tex.z-dn.net/?f=v_%7Boy%7D%C2%B2%20%2F2%20g%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y-%200%20%3D%2010.0%C2%B2%2F2%209.8%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y%20-%200%20%3D%205.10%20m%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%0A%3C%2Fp%3E%3Cp%3EThe%20height%20from%20the%20ground%20is%20the%20height%20that%20rises%20from%20the%20reference%20system%20plus%20the%20depth%20of%20the%20ground%20from%20the%20reference%20system%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%205.1%20%2B%2044%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%2049.1%20m%0A%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ELet%27s%20use%20the%20other%20equation%20to%20find%20the%20time%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20%20%5Btex%5Dv_%7By%7D)
= - g t
t = / g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 = ² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s
