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3 years ago

11

A ball is thrown horizontally from the top of a tall cliff. Neglecting air drag, what vertical distance will the ball have falle

n after 3 seconds?

1 answer:

adell [148]3 years ago

5 0

The relevant equation to use here is:

y = v0 t + 0.5 g t^2

where y is the vertical distance, v0 is initial velocity = 0, t is time, g = 9.8 m/s^2

y = 0 + 0.5 * 9.8 * 3^2

<span>y = 44.1 meters</span>

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5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m

iris [78.8K]

Answer:

<h2>470.59 kg</h2>

Explanation:

The the mass of the car can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{400}{0.85} \\ = 470.58823...$

We have the final answer as

<h3>470.59 kg</h3>

Hope this helps you

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3 years ago

Which statement accurately describes the molecules of a gas?

Triss [41]

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3 years ago

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

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3 years ago

You are out for a bicycle ride on a calm, windless day. You are heading northward on a level road and are experiencing a pressur

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The answer is (C)= zero but that it is swirling rapidly in all directions.

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3 years ago

a bullet is fired horizontally with a speed of 524 m/s from a height of 22m above the ground. calculate where it will hit the gr

lara31 [8.8K]

Given data:

* The height of the bullet is 22 m.

* The speed of bullet in the horizontal direction is 524 m/s.

Solution:

By the kinematics equation, the time taken by the bullet to reach the ground is,

$h=u_yt+\frac{1}{2}gt^2$

where u_y is the vertical velocity component, t is the time taken to reach the ground, g is the acceleration due to gravity, and h is the height of the bullet,

Substituting the known values,

$\begin{gathered} 22=0+\frac{1}{2}\times9.8\times t^2 \\ t^2=\frac{22\times2}{9.8} \\ t^2=4.49 \\ t=2.12\text{ s} \end{gathered}$

Thus, the time taken by the bullet to reach the ground is 2.12 seconds.

By the kinematics equation for the horizontal motion, the horizontal range of the bullet is,

$R=u_xt+\frac{1}{2}at^2$

where u_x is the horizontal component of the velocity, a is the acceleration along the horizontal direction, t is the time taken to reach the ground and R is the horizontal range,

Substituting the known values,

$\begin{gathered} R=524\times2.12+0 \\ R=1110.9\text{ m} \end{gathered}$

Thus, the horizontal range of the bullet is 1110.9 meters.

Hence, the bullet hit the ground at 1110.9 meters.

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1 year ago