Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
Answer:
The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is 
Explanation:
Given
x1 = 0 mm
x2 = 6 mm = 6 * 
m
c1 = 2 kg/
c2 = 0.4 kg/
T = 600 °C
Area = 0.25 
D = 1.7 * 
First equation
J = - D 
Second equation
J = 
To find the J (flux) use the First equation
J = - 1.7 * * 
To find M use the Second equation
= 
M =
Answer:
Eutectic product in Fe-C system is called Ledeburite-C.
Answer:
work done = 48.88 × 
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 × J
