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2 years ago

9

Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe

ed of 495 m/s in 1.78 s and can be brought to a stop again in 2.16 s.
Find the acceleration of the sled while accelerating.
Please be very thorough in your answer, I have a very hard time understanding if you don't.
Also, find the acceleration of the sled when braking.

1 answer:

Greeley [361]2 years ago

3 0

Answer:cho v₀ =0s

α=Δv/Δt

Explanation:

$\frac{0-495}{2,16-1,78}$

=-1302,631579

chuyển động chậm dầnđều

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A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

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3 years ago

Average velocity is different than average speed because calculating average velocity involves

Artemon [7]

Answer:

calculating displacement.

Explanation:

It's not true that displacement and distance would be the same always. Displacement is always smaller than or equal to distance as it is the smallest path between the initial and final point whereas distance is the measure of the total path covered.

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2 years ago

A cruise ship makes its way from one island to another. The ship is in motion compared with which reference point?

jekas [21]

A reference point would be something not on the ship which could be used to calculate distance traveled.

Answer: C.) A lighthouse on a nearby Island

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2 years ago

Read 2 more answers

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

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2 years ago

The surface of the Earth changes from processe such as erosion. Which of these changes to Earth's surface is an example of erosi

Ad libitum [116K]

Answer:

D. the wind picking up dust and carrying it

Explanation:

Erosion is a process in which an agent transfer the top soil to another region, thereby exposing the lower soil. These agents have the ability to move the top layer of soil and deposit it at another place. The major agents in this case are; a running or flowing body of water and wind.

Therefore, the change to the Earth's surface that is an example of erosion is the wind picking up dust and carrying it. Thereby exposing the lower layers.

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2 years ago