Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2
Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:
ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.
w=2*π*f
We know that the turntable is set to 33 1/3 rev/m so
the frequency 33.33/60=0.55 Hz
then w=2*π*0.55=3.45 rad/s
Finally the centripetal acceleration at differents radii results equal:
r= 0.05 m ac=3.45^2*0.05=0.50 m/s^2
r=0.1 ac=3.45^2*0.1=1.20 m/s^2
r=0.14 ac=3.45^2*0.14=1.66 m/s^2
Answer:
New volume, v2 = 0.8L
Explanation:
<u>Given the following data;</u>
Original Volume = 2L
Original Temperature = 280K
New Temperature = 700K
To find new volume V2, we would use Charles' law.
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles is given by;
Making V2 as the subject formula, we have;


V2 = 0.8L
Therefore, the volume of the gas after it is heated is 0.8L.
Answer:

Explanation:
We have,
The surface temperature of the star is 60,000 K
It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

Here,
= wavelength

So, the wavelength of the star is
.
Answer:
.
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is
(same as
,) meaning that
of energy is supplied to this lamp every second.
The question states that
out of that
of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the
of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive
of energy input and would outputs
of useful work. The efficiency of this lamp would be:
.
No mechanical waves cannot travel in a vacuum