Answer: 33 mm
Explanation:
Given
Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m
Internal pressure of gas, P(i) = 1.5 MPa
Yield strength of steel, P(y) = 340 MPa
Factor of safety = 0.3
Allowable stress = 340 * 0.3 = 102 MPa
σ = pr / 2t, where
σ = allowable stress
p = internal pressure
r = radius of the tank
t = minimum wall thickness
t = pr / 2σ
t = 1.5*10^6 * 4.5 / 2 * 102*10^6
t = 0.033 m
t = 33 mm
The minimum thickness of the wall required is therefore, 33 mm
Let's calculate the average acceleration. It is the rate of changing speeds. Hence, we need to calculate the difference of speeds. 10-6=4 m/s. The rate is now

.
In general, the formula for the mean acceleration between two times 1 and 2 is given by:

where v1 and v2 are the speeds at the respective points and T is the time interval between them.
Answer:
Increases
Explanation:
The kinetic energy _____ (Increses)
as the roller coaster goes downhill
Answer:
A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.
Explanation:
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.