So lunar eclips earth between sun and moon
Solar eclips moon between sun and earth.
About the 3th.. im not sure, it depends on if you meen a total solar eclips or not... i think total is more rare then a lunar eclipse..
Answer:
a) The final pressure is 1.68 atm.
b) The work done by the gas is 305.3 J.
Explanation:
a) The final pressure of an isothermal expansion is given by:
Where:
: is the initial pressure = 5.79 atm
: is the final pressure =?
: is the initial volume = 420 cm³
: is the final volume = 1450 cm³
n: is the number of moles of the gas
R: is the gas constant
Hence, the final pressure is 1.68 atm.
b) The work done by the isothermal expansion is:
Therefore, the work done by the gas is 305.3 J.
I hope it helps you!
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer: dakdadakdadakdadakda
Explanation:(sings) blah blah blah middle fingers in the air l-l-l-loser
Answer:
A. Zero
Explanation:
Given data,
The charge of the test charge, q = 1 C
The distance the charge moved against the filed of intensity, x = 30 cm
= 0.3 m
The electric field intensity, E = 50 N/C
The energy stored in the charge at 0.3 m is given by the formula,
V = k q/r
Where,
= 9 x 10⁹ Nm²C⁻²
The charge is moved from the potential V₁ to V₂ at 30 cm
Substituting the given values in the above equation
V₁ = 9 x 10⁹ x 30 / 0.3
= 1.5 x 10¹² J
And,
V₂ = 1.5 x 10¹² J
The energy stored in it is,
W = V₂ - V₁
= 0
Hence, the energy stored in the charge is, W = 0
