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Bezzdna [24]
3 years ago
7

Ball A is thrown vertically upwards with a velocity of v0 . Ball B is thrown upwards from the same point with the same velocity

t seconds later.
Part A

Determine the elapsed time t<2v0/g from the instant ball A is thrown to when the balls pass each other .

Express your answer in terms of some or all of the variables v0, t, and the acceleration due to gravity g .
Physics
1 answer:
Illusion [34]3 years ago
7 0

Answer:

The answer is \tau = \frac{v_o}{g} + \frac{t}{2}

Explanation:

If the ball is thrown vertically, the equation of its position is

y(\tau) = y_0 + v_0\tau - \frac{1}{2}g\tau^2

So setting our coordinate system in the position of throwing (y_0=0), the equation for A is

y_A(\tau) = v_0\tau - \frac{1}{2}g\tau^2

and for B

y_B(\tau') = v_0\tau' - \frac{1}{2}g{\tau'}^2

where

\tau' = \tau - t

due to the delay of the throwing between A and B, "t"

t = \tau - \tau'.

Now, the balls passing each other means that

y_B(\tau') = y_A(\tau)

then

v_0(\tau - t) - \frac{1}{2}g(\tau - t)^2 = v_0\tau - \frac{1}{2}g\tau^2

cancelling some terms...

-v_0t + g\tau t - \frac{1}{2}gt^2 = 0

so

\tau = \frac{v_o}{g} + \frac{t}{2}

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Soloha48 [4]

The speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

<h3>How to calculate speed of a wave?</h3>

The speed of a wave can be calculated by using the following formula:

Speed = Wavelength x Frequency

According to this question, Linh creates waves in the spring by moving the other end right and left with a frequency of 2 Hz. If wave crests are 0.5 m apart, the speed can be calculated as follows:

speed = 2Hz × 0.5m

speed = 1m/s

Therefore, the speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

Learn more about speed at: brainly.com/question/10715783

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7 0
2 years ago
Pls help i will be so happy thank you
Anvisha [2.4K]

Answer:

below given.

Explanation:

1) linear/direct

2) Indirect

3) linear/direct

4)  quadratic

5) Inverse

6) linear/direct

7) Inverse / Indirect

8) Inverse / Indirect

4 0
3 years ago
slader the cross section of a 5-ft long trough is an isosceles trapezoid with a 2 foot lower base, a 3-foot upper base, and an a
Ostrovityanka [42]

Answer:

0.08 ft/min

Explanation:

To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:

l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section <em>A</em> at that point:

A=d\times l\\A=5\,ft \times 2.5\, ft\\A=12.5\, ft^{2}

And the raising speed <em>v </em>of the water is given by:

v=\frac{q}{A}\\v=\frac{1\, \frac{ft^3}{min}}{12.5\, ft^2}\\v=0.08\, \frac{ft}{min}

where <em>q</em> is the water flow (1 cubic foot per minute).

7 0
3 years ago
30 POINTS!!! CAN U AWNSER IT?? :)
solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

F\Delta t = m \Delta v As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, \Delta v = 69.4 m/s;

45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg

7 0
2 years ago
The technology in solar panels allows us to convert the _________________ energy from the sun to __________________________ ener
Natalija [7]
The first blank: HEAT
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3 years ago
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