Answer:
The greater the amplitude the greater the energy.
(Think of a water wave - which carries greater energy a 1 ft wave or
a 10 ft wave)
<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>
Explanation:
Specific heat capacity
It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .
It is given as :
Heat absorbed = mass of substance x specific heat capacity x rise in temperature
or ,
Q= m x c x t
In above question , it is given :
For Q
mass of Q = m
Temperature changed =T₂/2
Heat supplied = x
Q= mc t
or
X=m x C₁ X T₁
or, X =m x C₁ x T₂/2
or, C₁=X x 2 /m x T₂ (equation 1 )
For another quantity : P
mass of P =m/2
Temperature= T₂
Heat supplied is same that is : X
so, X= m/2 x C₂ x T₂
or, C₂=2X/m. T₂ (equation 2 )
Now taking ratio of C₂ to c₁, We have
C₂/C₁= 2X /m.T₂ /2X /m.T₂
so, C₂/C₁= 1/1
so, the ratio is 1: 1
Answer:
S = 2 π R
R (mean) = 92.9E6 miles
S = 2 * 3.14 * 92.9E6 miles = 5.84E8 miles
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
