Answer:
R min = 28.173 ohm
R max = 1.55 × ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = ...........1
so R will be
R = ....................2
put here value
so for t min 11.5 μs
R =
R min = 28.173 ohm
and
for t max 6.33 ms
R max =
R max = 1.55 × ohm
The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ =
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε =
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer:
(Option B)
Explanation:
The absolute pressure of the air-filled tank is:
Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity
So, the potential energy of the mass is 2940 J.