The equilibrium temperature of aluminium and water is 33.2°C
We know that specific heat of aluminium is 0.9 J/gm-K, and that of water is 1 J/gm-K
Now we can calculate the equilibrium temperature
(mc∆T)_aluminium=(mc∆T)_water
15.7*0.9*(53.2-T)=32.5*1*(T-24.5)
T=33.2°C
Answer:
Explanation:
First, denote our known values;
Mass is impulse divided by change in velocity:
Hence, the mass of the ball is 141.30grams
Answer:
<u>We are given: </u>
initial velocity (u) = 0 m/s
final velocity (v) = 10 m/s
displacement (s) = 20 m
acceleration (a) = a m/s/s
<u>Solving for 'a'</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(10)² - (0)² = 2(a)(20)
100 = 40a
a = 100 / 40
a = 2.5 m/s²
Answer:
a) Initial angular speed = 30 rad/s
b) Final angular speed = 70 rad/s
Explanation:
a) We have equation of motion s = ut + 0.5at²
Here s = 400 radians
t = 8 s
a = 5 rad/s²
Substituting
400 = u x 8 + 0.5 x 5 x 8²
u = 30 rad/s
Initial angular speed = 30 rad/s
b) We have equation of motion v = u + at
Here u = 30 rad/s
t = 8 s
a = 5 rad/s²
Substituting
v = 30 + 5 x 8 = 70 rad/s
Final angular speed = 70 rad/s
