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Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s

Answers

Answer 1

Answer:

Answer:

The linear velocity is represented by the following expression: $v = (s)/(t)$

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

$v = r\cdot \omega$ (Eq. 1)

Where:

$r$ - Radius of rotation of the particle, measured in meters.

$\omega$ - Angular velocity, measured in radians per second.

$v$ - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

$\omega = (\theta)/(t)$ (Eq. 2)

Where:

$\theta$ - Angular displacement, measured in radians.

$t$ - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

$v = (r\cdot \theta)/(t)$ (Eq. 3)

From Geometry we must remember that circular arc ( $s$ ), measured in meters, is represented by:

$s = r\cdot \theta$

$v = (s)/(t)$

The linear velocity is represented by the following expression: $v = (s)/(t)$

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A piston above a liquid in a closed container has an area of 0.75m^2, and the piston carries a load of 200kg. What will be the external pressure on the upper surface of the liquid?

Answers

Answer:

2613.3 pa

Explanation:

p=F/A

p=ma/A

p=200×9.8/0.75

p=2613.3

According to the World Flying Disk Federation, the world distance record for a flying disk throw in the men’s 85-years-and-older category is held by Jack Roddick of Pennsylvania, who on July 13, 2007, at the age of 86, threw a flying disk for a distance of 54.0 m. If the flying disk was thrown horizontally with a speed of 13.0 m/s, how long did the flying disk remain aloft? (Jack Roddick was also a physics teacher! Read more about him at

Answers

Answer:

t = 4.15 seconds

Explanation:

It is given that,

Distance traveled by a flying disk, d = 54 m

The speed at which it was thrown, v = 13 m/s

We need to find the time for which the flying disk remain aloft. Let the distance is d. We know that, speed is equal to the distance covered divided by time. So,

$t=(d)/(v)\n\nt=(54\ m)/(13\ m/s)\n\nt=4.15\ s$

Hence, for 4.15 seconds the flying disk remain aloft.

Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x

Answers

Answer:

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

Explanation:

For this exercise let's start by applying Newton's second law to the mass with the string

W - T = m a

In this case, as the system is going down, we will assume the vertical directional down as positive.

T = W - m a

Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations

∑ τ = I α

T r = I α

the moment of inertia of the disk is

I = ½ M R²

angular and linear acceleration are related

a = α r

we substitute

T r = (½ m R²) (a / r)

T = ½ m ( $(R)/(r)$ )² a

we write our two equations

T = W - m a

T = ½ m ( $(R)/(r)$ )² a

we solve the system of equations

W - m a = ½ m (\frac{R}{r} )² a

m g = m a [ 1 + ½ (\frac{R}{r} )² ]

a = $(g)/( 1 + (1)/(2) ((R)/(r))^2 )$

this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations

w² = w₀² + 2 α θ

v² = v₀² + 2 a y

as the system is released its initial angular velocity is zero

w² = 0 + 2 α θ

v² = 0 + 2 a y

we look for the angular acceleration

a =α r

α = a / r

α = $(g)/(r (1 + (1)/(2) ((R)/(r))^2 )$

we look for the angle, remember that they must be measured in radians

θ = s / r

in this case we approximate the arc to the distance

s = y

θ = y / r

we substitute

w = $\sqrt{2 (g)/( r( (1)/(2) ((R)/(r))^2 ) (y)/(r) }$

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

for the simple case where r = R

w = $\sqrt{ (2gy)/( (3)/(2) R^2 ) }$

w = $\sqrt{ (4)/(3) (gy)/(R^2) }$

If the radio waves transmitted by a radio station have a frequency of 83.5 MHz, what is the wavelength of the waves, in meters

Answers

Answer: wavelength =3.52m

Explanation:

,λ=c/μ

where c=speed of the light,λ=wave length, μ=frequncy

c=3x10^8m/s

And

μ=83.5/MHz =85.3x10^6Hz==85.3x10^6Hz=

=85.3x10^6s-1

λ=c/μ

=3x10^8m/s/85.3x10^6s-1

=3.51699883

=3.52m

You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling fan with light kits be installed in five different rooms. Each fan contains a light kit that can accommodate four 60-watt lamps. Each fan motor draws a current of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than three hours at a time and therefore must be considered a continuous-duty device. The fans are to be connected to a 15-ampere circuit. Because the devices are continuous-duty, the circuit current must be limited to 80% of the continuous connected load. How many fans can be connected to a single 15-ampere circuit

Answers

Answer:

3 fans per 15 A circuit

Explanation:

From the question and the data given, the light load let fan would have been

(60 * 4)/120 = 240/120 = 2 A.

Next, we add the current of the fan motor to it, so,

2 A + 1.8 A = 3.8 A.

Since the devices are continuos duty and the circuit current must be limited to 80%, then the Breaker load max would be

0.8 * 15 A = 12 A.

Now, we can get the number if fans, which will be

12 A/ 3.8 A = 3.16 fans, or approximately, 3 fans per 15 A circuit.

Final answer:

The total power draw of each fan is 3.8 amperes. Thus, considering a limit of 80% usage of 15 amperes, only 3 fans can be connected to a single circuit to keep the total power draw below 12 amperes.

Explanation:

The question is asking how many ceiling fans, each with a certain power draw, can be connected on a single 15-ampere circuit, considering that each fan is a continuous-duty device. The power draw of each fan when the motor is operated at high speed and the light kit is fully loaded is the sum of the power draw of the motor and the light kit. As the power draw of each motor is 1.8 amperes and the light kit is 240 watts or 2 amperes (calculated using the formula Power = Voltage x Current; assuming a voltage of 120 volts), the total power draw of each fan is 3.8 amperes. Considering the limit of 80% of the continuous load, only 12 amperes (80% of 15) can be used. Thus, 3 fans can be connected to the circuit as it reaches 11.4 amperes, close enough to the 12 amperes limit.

Learn more about Circuit Capacity here:

brainly.com/question/34704571

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What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

Answers

Answer:

E = 9.4 10⁶ N / C, The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀ λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹² 3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

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