Answer:
The average velocity is 0.15 m/s
Explanation:
Use the definition of average velocity as the distance traveled divided the time it took.
Since the movement was on the plane from the origin (0, 0) to the point (-30, 20) corresponding to 30 m west and 20 m north, we calculate the distance using the distance between two points on the plane:
Then the magnitude of the average velocity can be estimated via the quotient between distance divided time, but since the units required are meters per second, we first convert the four minute time into seconds: 4 * 60 = 240 seconds.
Then the average velocity becomes:
A) visible light because it just makes since
To solve this problem we will apply the concepts related to the conservation of momentum. The momentum can be defined as the product between the mass of the object and its velocity, and the conservation of the momentum as the equality between the change of the initial momentum versus the final momentum. Mathematically, this relationship can be described as
Here,
= Mass of each object
= Initial velocity of each object
= Final velocity of each object
According to the statement one of the bodies does not have initial velocity, therefore said term would be zero. And the equation could be rewritten as,
Replacing the values respectively (The mass of your body with its respective speed we would have)
Therefore the initial velocity of the 2kg cart is 0.55m/s
Answer:
70,100 m/s
Explanation:
Given:
v₀ = 205 m/s
a = 8.03 m/s²
Δx = 1750 m
Find: v
v² = v₀² + 2aΔx
v² = (205 m/s)² + 2 (8.03 m/s²) (1750 m)
v = 70,130 m/s
Rounding to three significant figures, the final velocity is 70,100 m/s.
<span>Ans : Bernoulli's principle states for incompressible non-viscous flow that
p/Ď + gâ™h + (1/2)â™v² = constant
Evaluate the equation along a stream line from liquid surface of the reservoir (1) to the inlet of the pipe
pâ‚/Ď + gâ™hâ‚ + (1/2)â™v₲ = pâ‚‚/Ď + gâ™hâ‚‚ + (1/2)â™v₂²
=>
vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) + v₲ ]
lets make some assumptions:
- the pressure at the liquid surface is equal to the atmospheric pressure
pâ‚ = 1atm = 101325Pa
- the velocity of the liquid at the surface (that is the speed at which the liquid level in reservoir decreases) is quite small, so it may be ignored:
v₠≠0
So
vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) ]
The height difference is fixed. So the only variable remaining is the pressure in the pipe. As higher it is as lower the velocity in the pipe is.
So you get the maximum velocity for the minimum pressure.
Since pressure cannot drop below zero this is
pâ‚‚ = 0
Therefore
vâ‚‚max = âš[ pâ‚/Ď + gâ™(hâ‚-hâ‚‚) ]
= âš[ 2â™101325Pa/1000kg/mÂł + 2â™9.81m/s²â™12m ]
= 20.93m/s</span>
