Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball, 
After the collision, final momentum 
Using the conservation of momentum as :


v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules
Answer:
m = 4.29 kg
Explanation:
Given that,
Mass of the object, m = 2.8 kg
Stretching in the spring, x = 0.018 m
Frequency of vibration, f = 3 Hz
Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :



k = 1524.44 N/m
Since, 


m = 4.29 kg
So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.
Answer:
x = D (M/M-m) 2.41
Explanation:
a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation
F = g m₁m₂ / r²
Σ F = 0
F1- F2 = 0
F1 = F2
We set the reference system in the body of greatest mass (M) the planet
F1 = g m₁ M / x²
F2 = G m1 m / (D-x)²
G m₁ M / x² = G m₁ m / (D-x)²
M (D-x)² = m x²
MD² -2MD x + M x² = m x²
x² (M-m) -2MD x + MD² = 0
We solve the second degree equation
x = [2MD ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)
x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)
x = D {M ± Ra (2M²-mM)} / (M-m)
x = D (M ± M √ (2-m/M)) / (M-m)
x = D (M / (M-m)) (1 ±√ (2-m/M)
Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root
x = D (M / (M-m)) (1 + √2)
x = D (M/M-m) 2.41
b) X = 2/3 D
x = D (M/M-m) 2.41
2/3 D = D (M/(M-m)) 2.41
2/3 (M-m) = M 2.41
2/3 M - 2/3 m = 2.41 M
1.743 M = 0.667 m
M/m = 0.667/1.743
M/m = 0.38
Answer:
a)
, b)
, c) 
Explanation:
a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.

The mass flow rate is:

According to property water tables, specific enthalpies and entropies are:
State 1 - Superheated steam




State 2s - Liquid-Vapor Mixture




The isentropic efficiency is given by the following expression:

The real specific enthalpy at outlet is:



State 2 - Superheated Vapor




The mass flow rate is:


b) The temperature at the turbine exit is:

c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:



