The answer to your question is "20kgx9.8m/s" because weight is the force an object is exerting on another object, and the formula used to calculate force is <em>Force = Mass * Acceleration</em>.
Answer:
21.3 V, 1.2 A
Explanation:
1.
These resistors are in series, so the net resistance is:
R = R₁ + R₂ + R₃
R = 20 + 30 + 45
R = 95
So the current is:
V = IR
45 = I (95)
I = 9/19
So the voltage drop across R₃ is:
V = IR
V = (9/19) (45)
V ≈ 21.3 V
2.
First, we need to find the equivalent resistance of R₂ and R₃, which are in parallel:
1/R₂₃ = 1/R₂ + 1/R₃
1/R₂₃ = 1/10 + 1/10
R₂₃ = 5
Now we find the overall resistance by adding the resistors in series:
R = R₁ + R₂₃ + R₄
R = 10 + 5 + 10
R = 25
So the current through R₁ is:
V = IR
30 = I (25)
I = 1.2 A
Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
now, Pressure at depth x
integrating both side
now,
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
Answer:
0.0133 A
Explanation:
The time at which B=1.33 T is given by
1.33 = 0.38*t^3
t = (1.33/0.38)^(1/3) = 1.52 s
Using Faraday's Law, we have
emf = - dΦ/dt = - A dB/dt = - A d/dt ( 0.380 t^3 )
Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²
emf = - A*(3*0.38)*t^2
thus, the emf at t=1.52 s is
emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V
if the resistance is 0.390 ohms, then the current is given by
I = V/R = 0.0052/0.390 = 0.0133 A
