Thermal energy comes from the ,movement of particles which produces heat, the faster the movement is the more heat
1 horsepower is equal to 746 W, so the power of the engine is

The power is also defined as the energy E per unit of time t:

Where the energy corresponds to the work done by the engine, which is

. Re-arranging the formula, we can calculate the time t needed to do this amount of work:
Answer:
Robert Hooke used an early microscope to observe a cork sample. How did this help contribute to cell theory? It helped to show that cells contain water. ... It helped to show that some cells are visible to the naked eye.
Explanation:
If your teacher checks if it was copied just put it in your on words
Answer:
The tension in the two ropes are;
T1 = 23.37N T2 = 35.47N
Explanation:
Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity
W = 4.2×10 = 42N
The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.
Using triangular law of force and sine rule to get the tension we have;
If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;
T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}
T1/sin33.8° = T2/sin57.6° = 42/sin88.6°
From the equality;
T1/sin33.8° = 42/sin88.6°
T1 = sin33.8°×42/sin88.6°
T1 = 23.37N
To get T2,
T2/sin57.6°= 42/sin88.6°
T2 = sin57.6°×42/sin88.6°
T2 = 35.47N
Note: Check attachment for diagram.
Answer:
= 3,126 m / s
Explanation:
In a crash exercise the moment is conserved, for this a system formed by all the bodies before and after the crash is defined, so that the forces involved have been internalized.
the car has a mass of m = 1.50 kg a speed of v1 = 4.758 m / s and the mass of the train is M = 3.60 kg and its speed v2 = 2.45 m / s
Before the crash
p₀ = m v₁₀ + M v₂₀
After the inelastic shock
= m
+ M
p₀ = 
m v₀ + M v₂₀ = m
+ M
We cleared the end of the train
M
= m (v₁₀ - v1f) + M v₂₀
Let's calculate
3.60 v2f = 1.50 (2.15-4.75) + 3.60 2.45
= (-3.9 + 8.82) /3.60
= 1.36 m / s
As we can see, this speed is lower than the speed of the car, so the two bodies are joined
set speed must be
m v₁₀ + M v₂₀ = (m + M)
= (m v₁₀ + M v₂₀) / (m + M)
= (1.50 4.75 + 3.60 2.45) /(1.50 + 3.60)
= 3,126 m / s