The equation formula:
P V = n R T
1,245 * 2 l = n R * 300 K
n R = 1,245 * 2 : 300 = 8.3
P * 2.5 l = n R * 400 K
P * 2.5 = 8.3 * 400
P = 3,320 : 2.5 = 1328 J
Answer: A ) 1,328 joules.
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The total energy TE = mgh + 1/2 mU^2; where h = 20 m, g = 9.81 m/sec^2, and U = 10 mps. When the ball reaches max height H, all that TE will be potential energy PE = mgH = TE.
So there you are. TE = mgh + 1/2 mU^2 = mgH = TE from the conservation of energy. Solve for H.
1) H = (gh + 1/2 U^2)/g = h + U^2/2g = ? meters where everything on the RHS is given. You can do the math.
2) As the ball drops from H to h, it picks up KE as the potential energy mgH is converted when the potential energy is diminished to mgh, where h < H. So PE - pe = ke = mg(H - h) = 1/2 mv^2 so solve for v = sqrt(2g(H - h)) and, again, everything is given. You can do the math.
3) Same deal as 2) except now its V = sqrt(2gH) because all the PE = mgH = 1/2 mV^2 = KE when it is about to hit the ground. You can do the math.
Answer:
The sphere's volume charge density is 2.58 μC/m³.
Explanation:
Given that,
Radius of sphere R= 8.40 cm
Electric field 
Distance r= 16.8 cm
We need to calculate the sphere's volume charge density
Using Gauss's law
Put the value into the formula
Hence, The sphere's volume charge density is 2.58 μC/m³.
Answer: in this question, the only charge in the cavity is Q. Inside the conducting spherical shell, the electric field is zero.
While outside the shell, the electric field is given by: k(q + Q)/r²
Where;
K= is a constant which is given as, 8.99 x 10^9 N m² / C².
Q= source charge which creates the electric field
q= is the test charge which is used to measure the strength of the electric field at a given location.
r= is the radius
Explanation: Inside the conducting spherical shell, the electric field is zero since the Electric field vanishes everywhere inside the volume of a good conductor.
Pressure can be defined as the amount of force applied to a square unit area
