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3 years ago

On an amusement park ride passengers accelerate straight downward from rest to 22.9 m/s in 2.2 s.

Physics

2 answers:

Kamila [148]3 years ago

8 0

Given,

Initial Velocity (u) = 0 m/s

Final Velocity (v) = 25.9 m/s

Time (t) = 2.5 sec

avg. acceleration = ?

avg. acceleration = v-u/t

avg = 25.9/2.5 = 10.36 m/s²

avg = 10.36 m/s²

Sindrei [870]3 years ago

5 0

Answer:

Acceleration, $a=10.4\ m/s^2$

Explanation:

Given that,

Initial velocity of the ride, u = 0 (at rest)

Final speed of the ride, v = 22.9 m/s

Time taken, t = 2.2 s

Let a is the acceleration of the passengers on this ride. It can be calculated using the formula of the acceleration as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{22.9-0}{2.2}$

$a=10.4\ m/s^2$

So, the time required to bring the car to a stop is $10.4\ m/s^2$ . Hence, this is the required solution.

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An Object is located 16 cm in front of a converging lens with focal length 12 cm. To the right of the converging lens, separated

ExtremeBDS [4]

Answer:

image is located at 22.5cm to the left of diverging lens.

Explanation:

In this question we have given,

object distance from converging lens,u=-16cm

focal length of converging lens, $f_{c}=12cm$

focal length of diverging lens, $f_{d}=-10cm$

distance between converging and diverging lens=20cm

we have to find location of final image,v=?

Here we will first find the location of image formed by converging lens

we know that u, v and $f_{c}$ are related by following formula

$\frac{1}{f_{c}} =\frac{1}{v}- \frac{1}{u}$ .............(1)

put values of $f_{c}$ and u in equation (1)

we got,

$\frac{1}{12} =\frac{1}{v}- \frac{1}{-16}$

$\frac{1}{12}-\frac{1}{16} =\frac{1}{v}$

$\frac{16-12}{16\times 12} =\frac{1}{v}$

$v=\frac{16\times 12}{16-12}\\ v=48cm$

The image is located 48 cm to the right of the converging lens. This image is real and inverted.

This image will object for the diverging lens which is now located (48-20=18) cm at the right hand side of diverging lens

it means,

u=18cm

we know that u, v and $f_{d}$ are related by following formula

$\frac{1}{f_{d}} =\frac{1}{v}- \frac{1}{u}$ .............(2)

put values of $f_{d}$ and u in equation (2)

$\frac{1}{-10} =\frac{1}{v}- \frac{1}{18}$

$\frac{1}{-10}+\frac{1}{18} =\frac{1}{v}$

$\frac{-18+10}{10\times 18} =\frac{1}{v}$

$v=\frac{10\times 18}{10-18}\\ v=-22.5cm$

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Wewaii [24]

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To measure the deceleration of a car, without the value of the time of the movement, the <u>Torricelli equation</u> is used, which consists of:

${v_{f}}^{2}={v_{o}}^{2}+2a\Delta s$

Where is final velocity, is initial velocity, is acceleration, and is displacement.

Now, substitute the values in the formula:

$0^{2} = (3.5)^{2} + 2a6$

$0 = 12.25 + 12a$

$a = -\frac{12.25}{12}$

$a = -1.02$

Finally, the value of the acceleration found is multiplied by the mass of the object, thus measuring the friction force:

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Learn more about friction force at: brainly.com/question/13707283

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kakasveta [241]

Answer:

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This is the exact expression for weight change with height, but let's look at its values for some reasonable heights h = 6300 m (very high mountain)

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The correct answer is a

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