<span>The specific heat (or the amount of heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius) of copper is about 0.386 J/g/degree Celsius. This means that if we supply 0.386 J of energy to 1 gram of copper, its temperature will increase by 1 degree Celsius.</span>
7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
Answer:
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is 
Speed of nucleus at A is 
Potential at point A, 
Potential at point B, 
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
So, the speed at point B is 
.
Answer:
3.03e-19 J
Explanation:
Use the formula E = hc/λ
Where:
h (Planck's constant) = 6.626e-34 J*s
c (speed of light, constant) = 3.00e8 m/s
λ (wavelength) = 656e-9 m
E = energy (in Joules)
E = (6.626e-34 * 3.00e8) / 656e-9 = 3.03018293e-19 = 3.03e-19 J
Answer:
Electric Field intensity is zero.
The reason for that is:
All charges are placed at equal distances from the center of the square and have same magnitude and sign. This means they will exert equal and opposite forces on the test charge at the center. Net force will become Zero.
