Answer:
Potential energy plus kinetic energy equals mechanical energy because mechanical energy is basically just all of an object's energy, it's just two kinds of energy. The potential is stored inside and kinetic is being used. Both of those together is the total amount of the objects energy, which is the mechanical energy.
Explanation:
The final temperature of the seawater-deck system is 990°C.
<h3>What is heat?</h3>
The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.
The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.
Suppose for 1 kg of sea water, the heat transferred from the system is given by
3,710,000 = 1 x 3,930 x (T - 48.17)
T = 990°C to the nearest tenth.
The final temperature of the seawater-deck system is 990°C.
Learn more about heat.
brainly.com/question/13860901
#SPJ1
Answer:
It will cause kinetic energy to increase.
Explanation:
Given that Speed and Motion you went from the starting line to the finish line at different rates.
If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.
And since Kinetic energy K.E = 1/2mv^2
Increase in the mass of the body will definitely make the kinetic energy of the body to increase.
Since the time is the same, that means the speed V is the same.
Weight W = mg
m = W/g
The new kinetic energy will be:
K.E = 1/2(M + m)v^2
This means that there will be increase in kinetic energy.
Answer:
A hypothesis is a basically a theory proposed to a subject or refrence to an act with limited evidences.
Answer:
Part A) the angular acceleration is α= 44.347 rad/s²
Part B) the angular velocity is 195.13 rad/s
Part C) the angular velocity is 345.913 rad/s
Part D ) the time is t= 7.652 s
Explanation:
Part A) since angular acceleration is related with angular acceleration through:
α = a/R = 10.2 m/s² / 0.23 m = 44.347 rad/s²
Part B) since angular acceleration is related
since
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s
since
ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s
Part C) at t=0
v = v0 + a*(t-t0) = 51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s
ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s
Part D ) since the radial acceleration is related with the velocity through
ar = v² / R → v= √(R * ar) = √(0.23 m * 9.81 m/s²)= 1.5 m/s
therefore
v = v0 + a*(t-t0) → t =(v - v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s
t= 7.652 s
