Explanation:
Acceleration is the rate of change of velocity with time. When acceleration increases a body moves a faster velocity.
- In the graph acceleration at time t= 100s is rapidly increasing.
- At t = 20s, the acceleration of the body is getting started up.
A vehicle at time 100s will have a faster velocity compared to one at t = 20s
You need to observe the car at two different times.
-- The first time:
You write down the car's speed, and the direction it's pointing.
-- The second time:
You write down the car's speed and the direction it's pointing, again.
You take the data back to your lab to analyze it.
-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.
-- You compare the first and second direction. If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.
(Remember, "acceleration" doesn't mean "speeding up".
It means any change in speed or direction of motion.)
I think is True! Is the best answer. Because the trumpet it make them sounds like the lips with the musician and it vibrate.
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
Answer:
5.56 A
Explanation:
From the question,
Q = it.............. Equation 1
Where Q = charges, i = current, t = time.
Make i the subject of the equation
i = Q/t.............. Equation 2
Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds
Substitite these values into equation 2
i = 200/(0.6×60)
i = 5.56 A
Hence the magnitude of the current flowing through the circuit is 5.56 A
