Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
<em>A statement that is true for ALL of the examples of electromagnetic waves is that;</em>
A) They all move at the same speed in a vacuum
<u>The reason for qualifying 'in vacuum' is because EM waves of different frequencies often propagate at different speeds through material. Generally speaking, we say that light travels in waves, and all electromagnetic radiation travels at the same speed which is about 3.0 * 108 meters per second through a vacuum.</u>
Answer:
One year
Explanation:
In Florida, once you have had your learner's license for 1 year without any traffic convictions, you will receive an intermediate license.
The rule for Florida is that if a driver drives safely without any conviction during that year of his trial, he will receive an intermediate license. This means that he is now completely eligible and safe to drive around Florida's roads.
I hope the answer is helpful.
Thanks for asking.
Mass and distance
force /pull of gravity decreases with the increase in separation between the two bodies
the amount of gravity an object possesses is proportional to the mass of that object.
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s