To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.
By definition we know that the heat lost must be equal to the heat gained, ie
Where,
Q = Heat exchange
The heat exchange is defined as
Where,
Specific heat
m = mass
Change in Temperature
Therefore replacing we have that
Replacing with our values we have that
Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C
<span>Radiant energy travels in straight lines until it strikes an object where it can be absorbed, reflected or transmitted</span>
Answer:
mass multiplied by velocity (4 words but uh
Answer:
19.3
Explanation:
Assuming we have to find Specific gravity of gold.
As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object
so it is given by
specific gravity = weight of the object/weight of the water displaced
now we have
weight of the object = (density)(volume)g
weight of object = (19.3)(0.55)g
now weight of the liquid displaced is given by
weight of water displaced = (1 g/cm^3)(0.55ml)g
now we have
specific gravity = (19.3×0.55)/(1×0.55)
specific gravity= 19.3
Answer:
The final kinetic energy of the Helium nucleus (alpha particle) after been scattered through an angle of 120° is
8.00 x 10-13J
Explanation:
In Rutherford Scattering experiment, the collision of the helium nucleus with the gold nucleus is an ELASTIC COLLISION. This means that the kinetic energy is conserved ( The same before and after the collision).
Thus, the final kinetic energy of the helium nucleus is the same as initial kinetic energy (8.00 x 10^-13Joules)
Although, the kinetic energy is converted to potential energy in Coulomb's law equation.
That is,
1/2(mv^2) = (K* q1q2)/r
Where m is the mass of helium nucleus, v is its colliding velocity, k is electrostatic constant, q1 is the charge on helium nucleus, q2 is the charge on gold nucleus, r is impact parameter
