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2 years ago

A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. What is the acceleration of the ball in the vertical

Physics

1 answer:

blagie [28]2 years ago

6 0

The acceleration of the ball in the vertical direction is $9.8 m/s^2$ , downward.

Explanation:

This is a typical example of projectile motion, which consists of two independent motions:

- A uniform horizontal motion at constant velocity (since there are no forces in the horizontal direction)

- A vertical accelerated motion at constant acceleration (due to the presence of the force of gravity)

We are considering now the vertical motion only. There is only one force acting on the ball in this direction: the force of gravity, of magnitude

F = mg

where m is the mass of the ball and $g=9.8 m/s^2$ the acceleration due to gravity, downward. This means that the acceleration of the ball in this direction is (using Newton's second law)

$a=\frac{F}{m}=\frac{mg}{m}=g = 9.8 m/s^2$

Therefore, the acceleration of the ball in the vertical direction is $9.8 m/s^2$ , downward.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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What . Slides are placed upside down in a slide projector to correct for the _______ image produced. A. magnified B. inverted C.

Yuri [45]

The correct answer is B. Inverted image. This is because of all the lenses and light refractions and what not. The same things happens with our eyes except our brains fix the inverted image automatically. Since there are no brains in a projector, you have to fix it on your own by putting it in reverse.

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2 years ago

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An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

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2 years ago

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

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2 years ago

a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball

MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.

If it accelerates at 20 m/s² during the hit, then

Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = 4 newtons .

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3 years ago

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Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.

spin [16.1K]

Explanation:

The water cycle basically involves five steps:

evaporation and transpiration ⇄
condensation, ⇄
precipitation, ⇄
runoff, ⇄
infiltration ⇄

So when a thunderstorm occurs it helps in completing the precipitation process by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water runoffs to the surface of the ground, on plants, into rocks, rivers, and lakes.

Next, the Infiltration process enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the evaporation and transpiration process begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back into the atmosphere.

It then condenses and falls back as precipitation again.

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2 years ago