Answer:
Explanation:
Heat capacity A = 3 x heat capacity of B
initial temperature of A = 2 x initial temperature of B
TA = 2 TB
Let T be the final temperature of the system
Heat lost by A is equal to the heat gained by B
mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)
heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)
3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)
3 TA - 3 T = T - TB
6 TB + TB = 4 T
T = 1.75 TB
(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)
</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>
Answer:
B. 4 m/s
Explanation:
v=d/t
Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.
Answer:
remains the same, but the apparent brightness is decreased by a factor of four.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).
The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.
The apparent brightness of a star is measured in watts per square meter.
If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).
Some of the examples of stars are;
- Canopus.
- Sun (closest to the Earth)
- Betelgeuse.
- Antares.
- Vega.
Answer:
The correct answer is option 'd': The frequency decreases and the intensity of the sound decreases.
Explanation:
1) <u>Effect on Frequency </u>
According to Doppler's effect of sound we have
for a source of sound moving away from the observer the relation between the observed and the original frequency is given by
where
c = speed of sound in air
is the velocity of observer of sound
is the velocity of source of sound
is the original frequency of sound
As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.
2) <u>Effect on Intensity:</u>
At a distance 'r' from source emitting a wave of Power 'P' is given by
As we see on increasing 'r' intensity of sound decreases.
