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4 years ago

7

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

e motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 16 m , and you use your watch to find that each loop around takes 29 s. What is the speed?

1 answer:

Zepler [3.9K]4 years ago

5 0

Answer:

$v=3.47m/s$

Explanation:

The speed is by definition the distance traveled divided over the time it takes to travel that distance. In this case, this distance is the circumference of the wheel, so we have:

$v=\frac{C}{t}=\frac{2\pi r}{t}$

where we have written the circumference in terms of its radius.

For our values we then obtain the value:

$v=\frac{2\pi r}{t}=\frac{2\pi (16m)}{(29s)}=3.47m/s$

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Consider the program below:public class Test{ public static void main(String[] args) {Int[] a;a = new int[10];for(int i = 0; i &

ivolga24 [154]

Answer:

c. 65

Explanation:

The output is 65.

An array of length 10 is created first. Then, the first for-loop fill the array with different values; The array element now become: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The array element are generated using the equation a[i] = i + 2; so when i is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. i must be less than the array length (10).

a[0] = 0 + 2 = 2

a[1] = 1 + 2 = 3

a[2] = 2 + 2 = 4

a[3] = 3 + 2 = 5

a[4] = 4 + 2 = 6

a[5] = 5 + 2 = 7

a[6] = 6 + 2 = 8

a[7] = 7 + 2 = 9

a[8] = 8 + 2 = 10

a[9] = 9 + 2 = 11

result variable is declared and initialized to 0.

The second for-loop goes through the array and add individual element to result.

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3 years ago

A cell phone plan includes unlimited phone calls, texts, and data. There is an initial activation fee

Monica [59]

Answer:

90 months

Explanation:

1. 290 - 20 = 270

2. 270/3 = 90

4 0

3 years ago

Read 2 more answers

HURRY ! THE BEST I WILL GIVE THE BRAINLIEST

erik [133]

The speed of both cars is the same ... 80 km per hour.

But their velocities are different, because DIRECTION is part of velocity, and their directions are different.

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3 years ago

Pls help‼️‼️I have no idea how to solve this question... Can someone pls help! :(

sdas [7]

Answer: Its dragon ball bro

Explanation: (To Ship Leesoo) "My name is Twenty-Fifth Baam. For short you can just call me Baam. Even if you tell me to introduce myself, there's not much to say. I don't have parents... or a home... a single set of dirty clothes is all I have. I always thought that I might die alone without ever being able to do anything... But now- I have friends. Thank you."[45]

(Response when put in a danger of death situation) "There is something I fear more than death."[46][47]

(To Androssi Zahard) "The heart moves where the heart wills."[48]

(To Yu Han Sung) "If Rachel isn't up there, that place is nothing more than a pile of garbage. No matter what's up there, that won't change. You told me that being like I am is strange, but to me, you're a person who can't even find a single thing that's precious within that huge pile of garbage."[49]

(To Androssi Zahard) "If I must fight, I will fight to protect what is precious to me."[48]

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PART II

7 0

4 years ago

Read 2 more answers

A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) $I=0.0085\ A$

Explanation:

Given:

radius if the coil, $r=0.0395\ m$

no. of turns in the coil, $n=520$

variation of the magnetic field in the coil, $B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4$

resistor connected to the coil, $R=560\ \Omega$

(a)

we know, according to Faraday's Law:

$emf=n.\frac{d\phi}{dt}$

where:

$d \phi=$ change in associated magnetic flux

$\phi= B.A$

where:

A= area enclosed by the coil

Here

$A=\pi.r^2$

$A=\pi\times 0.0395^2$

$A=0.0049\ m^2$

$\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049$

So, emf:

$emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]$

$emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]$

$emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]$

$emf= 0.0306+3.516\times 10^{-4}\ t^3$

(b)

Given:

$t_0=5.25\ s$

Now, emf at given time:

$emf=4.7755\times 10^{-2}\ V$

∴Current

$I=\frac{emf}{R}$

$I=\frac{4.7755\times 10^{-2}}{560}$

$I=8.5\times 10^{-5} A$

6 0

4 years ago