Answer:
Δω = -5.4 rad/s
αav = -3.6 rad/s²
Explanation:
<u>Given</u>:
Initial angular velocity = ωi = 2.70 rad/s
Final angular velocity = ωf = -2.70 rad/s (negative sign is
due to the movement in opposite direction)
Change in time period = Δt = 1.50 s
<u>Required</u>:
Change in angular velocity = Δω = ?
Average angular acceleration = αav = ?
<u>Solution</u>:
<u>Angular velocity (Δω):</u>
Δω = ωf - ωi
Δω = -2.70 - 2.70
Δω = -5.4 rad/s.
<u> Average angular acceleration (αav):</u>
αav = Δω/Δt
αav = -5.4/1.50
αav = -3.6 rad/s²
Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.
Velocity is defined by rate of change in the position
which we can also write as

while acceleration is defined as rate of change in velocity

so acceleration and velocity both are rate of change in position and rate of change in velocity with respect to time respectively
out of all above statement the correct answer must be
<u>Acceleration equals change in velocity divided by time. </u>
Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 =<u> 1/9</u> 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
Answer : The heat change of the cold water in Joules is, 
Explanation :
First we have to calculate the mass of cold water.
As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.


Now we have to calculate the heat change of cold water.
Formula used :

where,
Q = heat change of cold water = ?
m = mass of cold water = 45 g
c = specific heat of water = 
= initial temperature of cold water = 
= final temperature = 
Now put all the given value in the above formula, we get:


Therefore, the heat change of cold water is 
Your teacher is right. The moon can be seen early in the morning sometimes and late at night. Different phases are only visible on certain days as one day might be full quarter, the next full moon, the next first quarter, etc.