Answer:x=2 and x=3
Explanation:
Given
Potential Energy for a certain mass is

and we know force is given by


For Force to be zero F=0




Therefore at x=2 and x=3 Force on particle is zero.
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
Answer:
12 N/cm²
Explanation:
From the question given above, the following data were obtained:
Weight (W) of block = 240 N
Area (A) = 20 cm²
Pressure (P) =?
Next, we shall determine the force exerted by the block. This can be obtained as follow:
Weight (W) of block = 240 N
Force (F) =.?
Weight and force has the same unit of measurement. Thus, we force applied is equivalent to the weight of the block. Thus,
Force (F) = Weight (W) of block = 240 N
Force (F) = 240 N
Finally, we shall determine the pressure on the floor as follow:
Force (F) = 240 N
Area (A) = 20 cm²
Pressure (P) =?
P = F/A
P = 240 / 20
P = 12 N/cm²
Therefore, the pressure on the floor is 12 N/cm².
Answer and Explanation:
curents i = 2.9 A
i ' = 4.4 A
the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed
= μo* ( i '- i )
Since from Ampere's law
where μ o = permeability of free space = 4π * 10 ^-7 H / m
plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)
= 1.884 * 10^-6 Tm
Answer:
Decreased by a factor of 4.5
Explanation:
"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

where
is the gravitational constant on Earth.
are the masses of the object and Earth itself. and R distance between, or the Earth radius.
So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:




Since
and 

So gravity would have been decreased by a factor of 4.5