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2 years ago

You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

Physics

2 answers:

Ksivusya [100]2 years ago

8 0

Answer:

19.6 m

Explanation:

Apex ; good luck to all :)

In-s [12.5K]2 years ago

4 0

-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.

As usual when we're discussing this stuff, we completely ignore air resistance.
</span>

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Light from a single laser is directed through two slits that are separated by a small distance. On the other side of the slits i

nekit [7.7K]

The answer is destructive interference. You have this for both C and D. I suspect one of C or D is supposed to be constructive interference... But destructive interference is the answer

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2 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

2 years ago

If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RC discharge thr

Agata [3.3K]

Answer:

Resistance in the flash tube, $R=3.97\times 10^{-3}\ \Omega$

Explanation:

It is given that,

Speed of the bullet, v = 500 m/s

Distance between one RC constant, d = 1 mm = 0.001 m

Capacitance, $C=503\ \mu F=503\times 10^{-6}\ F$

The time constant of RC circuit is given by :

$\tau=RC$

R is the resistance in the flash tube

$R=\dfrac{\tau}{C}$ ..........(1)

Speed of the bullet is given by total distance divided by total time taken as :

$v=\dfrac{d}{\tau}$

$\tau=\dfrac{d}{v}$

$\tau=\dfrac{0.001}{500}$

$\tau=0.000002\ s$

Equation (1) becomes :

$R=\dfrac{0.000002}{503\times 10^{-6}}$

$R=3.97\times 10^{-3}\ \Omega$

So, the resistance in the flash tube is $3.97\times 10^{-3}\ \Omega$ . Hence, this is the required solution.

8 0

2 years ago

Read 2 more answers

g A motorist traveling at 30.0 m/s passes a stationary police car. The police car gives chase 2.0 s later, accelerating at 5.0 m

Sedaia [141]

The time for the police car to catch up with the speeding motorist is 7.6 seconds.

<h3>What time will the police car catch up with the speeding motorist?</h3>

The police car and the motorist will cover equal distances.

Let the distance covered be d.

Distance covered by the motorist = speed * time

time = t, speed = 30 m/s

d = 30t

Distance covered by the police car = acceleration * (time)

time = t - 2, acceleration = 5.0 m/s²

d = 5(t-2)²

d = 5(t² - 4t + 4)

d = 5t² - 20t + 20

Equating the two equations for distance

5t² - 20t + 20 = 30t

5t² - 50t + 20 = 0

Solving for t using the quadratic formula:

t = 9.6 second or 0.4 seconds

Since t > 2, t = 9.6 seconds

t - 2 = 9.6 - 2

t - 2 = 7.6 seconds

Therefore, the time for the police car to catch up with the speeding motorist is 7.6 seconds.

Learn more about distance and acceleration at: brainly.com/question/14344386

#SPJ1

4 0

1 year ago

A horse is pulling 53.0 kg plow forward while the ground exerts a 275 N force, and the plow accelerates at 0.222 m/s^2. what is

JulsSmile [24]

Answer:
Magnitude of force the ground exerts on the plow = 263.234 N

Explanation:
Magnitude of force the ground exerts on the plow = Fground - Fplow
We are given that:
Fgound = 275 N
We will now calculate Fplow as follows:
Fplow = mass of horse * acceleration of plow
Fplow = 53 * 0.222
Fplow = 11.766 N

Now, substitute in the above equation to get magnitude of force the ground exerts on the plow as follows:
Magnitude of force the ground exerts on the plow = Fground - Fplow
Magnitude of force the ground exerts on the plow = 275 - 11.766
Magnitude of force the ground exerts on the plow = 263.234 N

Hope this helps :)

3 0

2 years ago