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
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You can estimate one more digit past the smallest division on the measuring device. If you look at a 10mL graduated cylinder, for example, the smallest graduation is tenth of a milliliter (0.1mL). That means when you read the volume, you can estimate to the hundredths place (0.01mL).
Answer: Hello!
Lewis is travelling at 165 mph, which means miles per hour, this says that he does 165 miles in one hour.
We want to know how much time takes to cover 16 miles.
this can be calculated as the quotient of the distance and the velocity; this is:

if we want to write this in minutes, then:
we know that one hour has 60 minutes, then 0.096 hours has:
0.096h*60mins/1h = 5.8 minutes.
then Lewis needs 5.8 minutes in order to cover 16 miles if his speed is 156 miles per hour.
Answer:

Explanation:
As we know that box was initially at rest
so here work done by all forces on the box = change in its kinetic energy
so we will have

now we have
m = 10 kg

L = 3.5 m
so we will have


so final speed is given as

Answer:
a. 2.0secs
b. 20.4m
c. 4.0secs
d. 141.2m
e. 40m/s, ∅= -30°
Explanation:
The following Data are giving
Initial speed U=40m/s
angle of elevation,∅=30°
a. the expression for the time to attain the maximum height is expressed as

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

b. the expression for the maximum height is expressed as

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,
Hence T=2t
T=2*2.0
T=4.0secs
d. The range of the projectile is expressed as

e. The landing speed is the same as the initial projected speed but in opposite direction
Hence the landing speed is 40m/s at angle of -30°