Someone help me with this ASAP!!

A Ferris wheel with 60 spokes has a diameter of 100 m. It makes one rotation every 60 seconds. Find the speed of the passengers

Doss [256]

Answer:

<em>The speed of the passengers is 5.24 m/s</em>

Explanation:

<u>Uniform Circular Motion </u>

It occurs when an object in a circular path travels equal angles in equal times.

The angular speed can be calculated in two different ways:

$\displaystyle \omega=\frac{v}{r}$

Where:

v = tangential speed

r = radius of the circle described by the rotating object

Also:

$\omega=2\pi f$

Where:

f = frequency

Since the frequency is calculated when the number of revolutions n and the time t are known:

$\displaystyle f=\frac{n}{t}$

The Ferris wheel has a diameter of 100 m and makes n=1 rotation in t=60 seconds, thus the frequency is:

$\displaystyle f=\frac{1}{60}\ Hz$

The angular speed is:

$\displaystyle \omega=2\pi \frac{1}{60} =\frac{\pi}{30} \ rad/s$

Now we calculate the tangential speed, solving this formula for v:

$\displaystyle \omega=\frac{v}{r}$

$v=\omega . r$

The radius is half the diameter, r=100/2=50 m:

$\displaystyle v=\frac{\pi}{30} . 50$

Calculating:

v = 5.24 m/s

The speed of the passengers is 5.24 m/s

7 0

3 years ago

Calcula la fuerza magnética de una carga que se desplaza hacia la derecha con una velocidad de 5 X 10 a la 6 metros sobre segund

Diano4ka-milaya [45]

Answer:

La fuerza magnética del electrón es 6.41x10⁻¹³ N.

Explanation:

Podemos encontrar la fuerza magnética (F) usando la ley de Lorentz:

$F = qv\times B$

En donde:

q: es la carga del electrón = 1.602x10⁻¹⁹ C

v: es la velocidad del electrón = 5x10⁶ m/s

B: es el campo magnético = 0.8 T

$F = qv\times B = 1.602 \cdot 10^{-19} C*5\cdot 10^{6} m/s*0.8 T = 6.41 \cdot 10^{-13} N$

Por lo tanto, la fuerza magnética del electrón es 6.41x10⁻¹³ N.

Espero que te sea de utilidad!

3 0

2 years ago

A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque

labwork [276]

Answer:

specific gravity = 0.8

specific gravity of solution = 2

Explanation:

given data

rectangular block above water = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block that is

specific gravity = block vol below ÷ total block vol × specific gravity water ..............1

put here value we get

specific gravity = $\frac{1.60}{1.60+0.400}$ × 1

specific gravity = 0.8

and now we get here specific gravity of solution that is express as

specific gravity of solution = total block vol ÷ block vol below × specific gravity block ........................2

put here value we get

specific gravity of solution = $\frac{1.60+0.400}{0.800}$ × 0.8

specific gravity of solution = 2

6 0

3 years ago

4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o

goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

(i) 10% of the peak output and (ii) 1% of the peak output. In each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0

3 years ago

During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck

shutvik [7]

<h2>Answer:2.65 seconds</h2>

Explanation:

Let $a$ be the acceleration.

Let $u$ be the initial velocity.

Let $v$ be the final velocity.

Let $t$ be the time taken.

As we know from the equations of motion,

$v=u+at$

Given, $v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}$

$0=8.12-3.06t$

$t=2.65sec$

6 0

3 years ago