Answer:
P = 17.28*10⁶ N
Explanation:
Given
L = 250 mm = 0.25 m
a = 0.54 m
b = 0.40 m
E = 95 GPa = 95*10⁹ Pa
σmax = 80 MPa = 80*10⁶ Pa
ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m
We get A as follows:
A = a*b = (0.54 m)*(0.40 m) = 0.216 m²
then, we apply the formula
ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L
⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)
⇒ P = 24624000 N = 24.624*10⁶ N
Now we can use the equation
σ = P/A
⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa
So σ > σmax we use σmax
⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N
Answer:
Yes the body will receive a dangerous shock in both cases.
Explanation:
Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.
Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.
Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:
I = V/R= 120/ 500*10^3
I= 0.24 mA
Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.
Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:
I = 120 / 1000
I = 12 mA
The current is higher than safe zone so the body will receive a dangerous shock.
The three types are alpha beta and gamma
Answer:
BOTH the size of the force AND the mass of the object
Explanation:
Acceleration of an object is the rate of change of its velocity.
The relation between force, mass and acceleration is given by the formula as follows :
F = ma
m is mass
a is acceleration
It would mean that the change in motion or the acceleration of an object depends on both the size of the force and the mass of the object. Hence, the correct option is (c).
Answer:
The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰
Explanation:
Given;
rock concert sound intensity level, β₁ = 120 dB
whisper sound intensity level, β₂ = 20 dB
The sound intensity level is given as;
where;
I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²
I is the sound intensity
Intensity of sound at rock concert ;
The intensity of sound of a whisper;
Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper
Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰