A) experimental because he isn’t sure and is testing out
Answer:
35.14°C
Explanation:
The equation for linear thermal expansion is
, which means that a bar of length
with a thermal expansion coefficient
under a temperature variation
will experiment a length variation
.
We have then
= 0.481 foot,
= 1671 feet and
= 0.000013 per centigrade degree (this is just the linear thermal expansion of steel that you must find in a table), which means from the equation for linear thermal expansion that we have a
= 22.14°. As said before, these degrees are centigrades (Celsius or Kelvin, it does not matter since it is only a variation), and the foot units cancel on the equation, showing no further conversion was needed.
Since our temperature on a cool spring day was 13.0°C, our new temperature must be
= 35.14°C
1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches
(55 gal) x (231 in³/gal) x (1 ft³/1728 in³)
= (55 x 231 / 1728) ft³
= 7.352 cubic feet (rounded)
Answer:

Explanation:
From the question we are told that
Mass 
Height 
Generally the equation for velocity before impact is is is mathematically given by



Generally the equation for Kinetic Energy is is mathematically given by




Explanation:
Given that,
Wavelength of the light, 
Work function of sodium, 
The kinetic energy of the ejected electron in terms of work function is given by :

The formula of kinetic energy is given by :

Hence, this is the required solution.