Answer:
87.1 mph
Explanation:
We are given that
Mass,m=60 kg
Power,P=340 W
Speed,v=5 m/s
Area,
Drag coefficient,
Coefficient of rolling resistance,
Friction force,
Where 
Let speed of cyclist=v'
Drag force,
Density of air,
Power,P=
1 m=0.00062137 miles
1 hour=3600 s
The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
Answer:
Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,
- separation distance of capacitor 1,
- quantity of charge on capacitor 2,
- quantity of charge on capacitor 1,
We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air
From eq. (1)
For capacitor 2:
For capacitor 1:
![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
& for capacitor 1:
![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)
Answer: atmospheric is air by the earth and pressure is just someone or something doing it
Explanation:
935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps
