Answer:
Explanation:
Given
Mass = 10kg
Velocity = 2m/s
Required
Calculate the momentum of the man
Momentum is calculated as thus
or
So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;
The formula becomes
Hence, the momentum of the man is 
The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .
If an object starts from rest and its speed after 2 seconds is 20 m/s, then its acceleration is 20/2 = 10 m/s^2 .
(Gravity on Earth is only 9.8 m/s^2.)
The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s
<h3>What are the components of velocity?</h3>
We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.
Hence;
Vertical component = 40 m/s sin 15 degrees = 10.35 m/s
Horizontal component = 40 cos 15 degrees = 38.6 m/s
Learn more about components of velocity:brainly.com/question/14478315
#SPJ1
The thermal energy that is generated due to friction is 344J.
<h3>What is the thermal energy?</h3>
Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.
Thus, the kinetic energy is given as;
KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J
PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J
The thermal energy is; 352.8 J - 9.1 J = 344J
Learn more about thermal energy due to friction:brainly.com/question/7207509
#SPJ1
Answer:
0° C
Explanation:
Given that
Mass of ice, m = 50g
Mass of water, m(w) = 50g
Temperature of ice, T(i) = 0° C
Temperature of water, T(w) = 80° C
Also, it is known that
Specific heat of water, c = 1 cal/g/°C
Latent heat of ice, L(w) = 89 cal/g
Let us assume T to be the final temperature of mixture.
This makes the energy balance equation:
Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C
m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have
50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)
4000 + 50T = 4000 - 50T
0 = 100 T
T = 0° C
Thus, the final temperature is 0° C
