Answer:
- the Thevenin equivalent circuit for the battery is uploaded below
- the battery delivered 117 watts of power to the starter motor
Explanation:
Given the data in the equation
diagram of the Thevenin equivalent circuit for the battery is uploaded below.
Current I = 10 A
Voltage 1 = 12 V
voltage 2 = 11.7 v
R = (V1 - V2) / I
R = (12-11.7)/10
R = 0.3 / 10
R = 0.03Ω
Thevenin equivalent circuit
= V2 / I = 11.7 / 10
= 1.17Ω
so, power delivered to the starter motor will be;
p = (V2)² /
P = ( 11.7 V )² / 1.17Ω
p = 136.89 / 1.17
p = 117 watts
Therefore, the battery delivered 117 watts of power to the starter motor
Answer:
a) μ = 0.1957
, b) ΔK = 158.8 J
, c) K = 0.683 J
Explanation:
We must solve this problem in parts, one for the collision and the other with the conservation of energy
Let's find the speed of the wood block after the crash
Initial moment. Before the crash
p₀ = m v₁₀ + M v₂₀
Final moment. Right after the crash
pf = m 
+ M v_{2f}
The system is made up of the block and the bullet, so the moment is preserved
p₀ = pf
m v₁₀ = m v_{1f} + M v_{2f}
v_{2f} = m (v₁₀ - v_{1f}) / M
v_{2f} = 4.5 10-3 (400 - 190) /0.65
v_{2f} = 1.45 m / s
Now we can use the energy work theorem for the wood block
Starting point
Em₀ = K = ½ m v2f2
Final point
Emf = 0
W = ΔEm
- fr x = 0 - ½ m v₂₂2f2
The friction force is
fr = μN
With Newton's second law
N- W = 0
N = Mg
We substitute
-μ Mg x = - ½ M v2f2
μ = ½ v2f2 / gx
Let's calculate
μ = ½ 1.41 2 / 9.8 0.72
μ = 0.1957
b) let's look for the initial and final kinetic energy
K₀ = 1/2 m v₁²
K₀ = ½ 4.50 10⁻³ 400²
K₀ = 2.40 10² J
Kf = ½ 4.50 10⁻³ 190²
Kf = 8.12 10¹ J
Energy reduction is
K₀ - Kf = 2.40 10²- 8.12 10¹
ΔK = 158.8 J
c) kinetic energy
K = ½ M v²
K = ½ 0.650 1.45²
K = 0.683 J
Answer:
we know that
ke=1÷2×mv^2
and when speed become triple of original then new speed become 3v
then,new ke=1÷2×m(3v)^2
=1÷2×9mv^2
9×original ke
hence ke will increase by 9 times
Answer:
LHS=RHS=[L]
Explanation:
Given mathematical expression:
where: dimension:
s = displacement length ![[L]](https://tex.z-dn.net/?f=%5BL%5D)
u = initial velocity ![[L.T^{-1}]](https://tex.z-dn.net/?f=%5BL.T%5E%7B-1%7D%5D)
t = time ![[T]](https://tex.z-dn.net/?f=%5BT%5D)
a = acceleration ![[L.T^{-2}]](https://tex.z-dn.net/?f=%5BL.T%5E%7B-2%7D%5D)
now using dimensional analysis:
![[L]=[L.T^{-1}]\times [T]+[L.T^{-2}] [T]^2](https://tex.z-dn.net/?f=%5BL%5D%3D%5BL.T%5E%7B-1%7D%5D%5Ctimes%20%5BT%5D%2B%5BL.T%5E%7B-2%7D%5D%20%5BT%5D%5E2)
we know that the ratio and constants have no dimension.
![[L]=[L]+[L]](https://tex.z-dn.net/?f=%5BL%5D%3D%5BL%5D%2B%5BL%5D)
as we know that only similar dimensions can be added or subtracted therefore we get a correct conclusion.
<em>However we can deduce the operators between the equations and can neither check for the validity of the constants. We can only check for the dimension of the terms involved.</em>
