Answer:
<em>The momentum of the car is 35,000 kg.m/s</em>
Explanation:
<u>Momentum</u>
Momentum is often defined as <em>mass in motion.</em>
Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:
If only magnitudes are considered:
p = mv
The car has a mass of m=1,000 kg and travels at v=35 m/s. Calculating its momentum:
p = 1,000 kg * 35 m/s
p = 35,000 kg.m/s
The momentum of the car is 35,000 kg.m/s
Answer:
0.4778 m/s
Explanation:
To solve this question, we will make use of law of conservation of momentum.
We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;
V_x = (12 m/s)(cos(35°)) = 9.83 m/s.
Thus, the horizontal component of the rock's momentum is;
(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.
Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.
Thus, to get the person's speed, we know that; momentum = mass x velocity
Mass of person = 72 kg and we have momentum as 34.405 kg·m/s
Thus;
34.405 = 72 x velocity
Velocity = 34.405/72
Velocity = 0.4778 m/s
Answer:
a) 2.85 kW
b) $ 432
c) $ 76.95
Explanation:
Average price of electricity = 1 $/40 MJ
Q = 20 kW
Heat energy production = 20.0 KJ/s
Coefficient of performance, K = 7
also
K=(QH)/Win
Now,
Coefficient of Performance, K = (QH)/Win = (QH)/P(in) = 20/P(in) = 7
where
P(in) is the input power
Thus,
P(in) = 20/7 = 2.85 kW
b) Cost = Energy consumed × charges
Cost = ($1/40000kWh) × (16kW × 300 × 3600s)
cost = $ 432
c) cost = (1$/40000kWh) × (2.85 kW × 200 × 3600s) = $76.95
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
There's no digram because I'm mr lemonade mr French fries
