Mass always stays the same therefore it can’t be c and d. Volume stays the same because she is not removing anything. Therefore it is a
An occluded front forms when a cold front catches up with a warm front. So it would bring Sun and warmth.
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:
The process is given in the pic.
I have taken the average masses so u substitute the values and solve hope it will help :)❤
Explanation:
We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.
Condition 1: The stars rise and set perpendicular to the horizon
The observer is at the equator
Condition 2: The stars circle the sky parallel to the horizon
The observer is at the Pole of the Earth
Condition 3: The celestial equator passes through the zenith
The observer is at the equator
Condition 4: In the course of a year, all stars are visible
The observer is at the equator
Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)
The observer is at North Pole
