<span>The question is what do all types of waves transfer. Waves, both mechanical and electromagnetic are disturbances that transfer energy without transporting matter. It doesnt matter wether they are propagating trough a medium or trough vacuum. Therefore the correct answer is A. energy.</span>
Answer:
Elastic Potential Energy
Explanation:
Elastic Potential Energy (“Spring Energy”) is the form of energy an object has when it is stretched, compressed, twisted, bent, or otherwise has its shape changed as long as the object resists and will try to return to its original state.
Answer:
The answer to this question is given below in this explanation section.
Explanation:
" law of conservation of energy"
The law of conservation of energy states that energy can neither be created nor destroyed only converted from one form of energy into another.This mean that a system always has a same account of a energy,unless it is added from the outside.This is particularly confusing in the case of non conversation forces,where energy is converted from ,mechanical energy into thermal energy.but the overall energy does remain the same.The only way to use energy is to transform energy from one form to another.
The amount of energy in any system than it is determined by the following equation.
Ut=Ui +W+Q
- Ut is the total internal energy of a system.
- Ui is the initial internal energy of a system.
- W is the work done by or on the system.
- Q is the heat added to or removed by the system.
It is also possible to determined the change in internal energy of the system using the equation.
ΔU=W+Q
The mechanical energy of a system increases provided their is no loss of energy due to friction.The energy would transform to kinetic energy when the speed is increasing.Te mechanical energy of a system remain constant provided their is no loss of energy due to friction.
The law of conversation of energy which say that in a closed system total energy is conserved that is it constant.
KE1 + PE1=KE2+PE2
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer:
a) The magnitude of the force is 968 N
b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N
Explanation:
<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>
Information given:
d = 106 km = 106,000 m
v1 = 28 m/s
G = 1.9 gal
η = 0.3
Eff = 1.2 x 10^8 J/gal
a) We can express the energy used as the work done. This work has the following expression:
Then, we can derive the magnitude of the force as:
b) We will calculate the force for a speed of 30 m/s.
If the force is proportional to the speed, we have:
