Answer:
The atomic model, the structure of the solar system, and the theory of black holes are examples of models created using indirect evidence.
Explanation:
I have no explanation sorry.
Answer:
Explanation:
Given that,
Spring constant = 16N/m
Extension of spring
x = 8cm = 0.08m
Mass
m = 5g =5/1000 = 0.005 kg
The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.
So, Using conservation of energy
Energy in spring is converted to kinectic energy
So, Ux = K.E
Ux = ½ kx²
Then,
Ux = ½ × 16 × 0.08m²
Ux = 0.64 J
Since, K.E = Ux
K.E = 0.64 J
Answer:
The increase in the internal energy = 350 J
Explanation:
Given that
Q= 275 J
W= - 125 J
W' = 50 J
W(net)= -125 + 50 = -75 J
Sign -
1.Heat rejected by system - negative
2.Heat gain by system - Positive
3.Work done by system = Positive
4.Work done on the system-Negative
Lets take change in the internal energy =ΔU
We know that
Q= ΔU + W(net)
275 = ΔU -75
ΔU= 275 + 75 J
ΔU=350 J
The increase in the internal energy = 350 J
Answer:
a) When its length is 23 cm, the elastic potential energy of the spring is
0.18 J
b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Explanation:
Hi there!
a) The elastic potential energy (EPE) is calculated using the following equation:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretched lenght.
Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).
First, let´s convert the spring constant units into N/m:
4 N/cm · 100 cm/m = 400 N/m
EPE = 1/2 · 400 N/m · (0.03 m)²
EPE = 0.18 J
When its length is 23 cm, the elastic potential energy of the spring is 0.18 J
b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:
EPE = 1/2 · 400 N/m · (0.06 m)²
EPE = 0.72 J
When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J
Answer:
0.01154 A
Explanation:
We have given the energy in the magnetic field
Value of inductance L =0.060 H
Energy stored in magnetic field is given by 


So the current flowing through rectangular toroid will be 0.01154 A