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Kamila [148]
2 years ago
7

A student wants to investigate the motion of a ball by conducting two different experiments, as shown in Figure 1 and Figure 2 a

bove. In Experiment 1, the student releases a ball from rest and uses a slow-motion camera to film the ball as it falls to the ground. Using video analysis, the student is able to plot the ball’s horizontal position x and vertical position y as a function of time t. In Experiment 2, the student horizontally rolls the same ball off a table, and uses video analysis to plot the ball’s horizontal position x and vertical position y as a function of time t starting from the instant the ball leaves the table. The graphs from each experiment are shown above along with each graphs best-fit curve line. In Experiment 1, what is the speed of the ball the instant it makes contact with the ground?
Physics
1 answer:
Lelu [443]2 years ago
5 0
It equals to G



^the answer
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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0
2 years ago
Joe learned that decaying plants produce an acidic environment. Joe wants to know if pill bugs prefer acidic environments instea
Anna11 [10]
Yessgshhshe the answer
6 0
2 years ago
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You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
2 years ago
A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

Initial speed (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

4 0
2 years ago
Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp
Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

Or

6 + 9 = 15 km

2 + 1 = 3h

15 + 3 = 18

15 x 2 = 30

3 x 2 = 6

30 - 6 = <u>24km/h</u>

8 0
2 years ago
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