It is fine to use the equation given by Plitter, since we are told that the mass is about the same as it is now, and I seriously doubt the original question wants the student to go into relativistic effects, electron degeneracy pressure and magnetic effects that govern a real white dwarf star.
There is no need to make it unnecessarily complicated, when the question is set at high school level. The question asks, given a particular radius, and a given mass, what will the density be (which in this case will be the average density). To answer the question, one needs to know the mass of the sun (which is about 2×1030 Kg. One needs to convert the diameter to a radius, and then calculate the spherical volume of the white dwarf. Then one can use the formula given above, namely density=mass/volume
im not 100 percent sure but i think its a
Answer: Mass and kinetic energy have a positive relationship, which means that as mass increases, kinetic energy increases, if all other factors are held constant
Explanation:
Answer:
1.37 seconds before.... splat !
Explanation:
y = y0 + v0 t + 1/2 a t^2 y0 = 30 y = 0
v0 = 0 ( from rest)
a = - 32.2 ft/s^2
0 = 30 - 1/2 * 32.2 * t^2
t=1.366 seconds <u> Note that mass is irrelevant.</u>
Answer:
a. 2.53 μJ b. It will move away
Explanation:
a. What is the electric potential energy between the particles?
The electric potential energy U = kq₁q₂/r where k = 9 × 10⁹ Nm²/C², q₁ = 4.1 nC = 4.1 × 10⁻⁹ C, q₂ = 2.4 nC = 2.4 × 10⁻⁹ C and r = distance between charges = 3.5 cm = 3.5 × 10⁻² m.
Substituting the values of the variables into U, we have
U = kq₁q₂/r²
U = 9 × 10⁹ Nm²/C² × 4.1 × 10⁻⁹ C × 2.4 × 10⁻⁹ C/3.5 × 10⁻² m
U = 88.56 × 10⁻⁹ Nm²/3.5 × 10⁻² m
U = 25.3 × 10⁻⁷ Nm
U = 2.53 × 10⁻⁶ Nm
U = 2.53 × 10⁻⁶ J
U = 2.53 μJ
b. And if the second particle is released will it move forward or away from the source charge.
It will move away from the source charge since they have the same sign of charge. Since, opposite charges repel.
