The answer to your question is A
The kinetic energy with which the hammer strikes the ground
is exactly the potential energy it had at the height from which it fell.
Potential energy is (mass) x (gravity) x (height) .... directly proportional
to height.
Starting from double the height, it starts with double the potential
energy, and it reaches the bottom with double the kinetic energy.
Answer:
, where the minus indicates the direction is opposite to that of the throw.
Explanation:
a)
Since MKS stands for meter-kilogram-second and we know that:
We can write that:
These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.
So we have that:
b)
Newton's 2nd Law tells us that F=ma, and the definition of acceleration is 
, so we have:
Taking the throw direction as the positive one, for our values we have:
Answer:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be 
Let the exit pressure be 
Ratio of reservoir pressure and exit pressure
= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
