What is the revolution ratio of two spur gears with one having 12 teeth and the other having 36 teeth

Answers

Answer 1

Answer: 3:1 is the ratio i hope this belps

Related Questions

Which sling can the crane use to lift the 1000kg pipe?A.800kg rated slingB. 1000kg rated slingC. 2000kg rated slingD. Band C
Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?
In an evironmental system of subsystem, the mass balance equation is:__________.
A student goes skateboarding a few times a week what is the dependent variable
The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom rises from pin A to B, at angle phi from horizontal. A cable goes up from the load to B and then left to a pulley. Determine the greatest weight of the crate that can be supported without causing the cable to fail if ϕ=30∘. Neglect the size of the winch.

A 30 kg child on a 2 m long swing is released from rest when the swing supports make an angle of 34 ◦ with the vertical. The acceleration of gravity is 9.8 m/s 2 . If the speed of the child at the lowest position is 2.31547 m/s, what is the mechanical energy dissipated by the various resistive

Answers

Answer:

Energy dissipated = 13.453 Joules

Explanation:

In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.

The gravitational potential energy (relative to lowest position) is found as follows:

G.P.E = mass * gravity * height

Where, Height = 2 - 2 * Cos(34°)

Height = 0.3193 meters

G.P.E = 30 * 9.8 * 0.3193

G.P.E = 93.874 J

Kinetic energy:

K.E = 0.5 * mass * velocity^2

K.E = 0.5 * 30 * 2.31547^2

K.E = 80.421 J

Energy dissipated = G.P.E - K.E

Energy dissipated = 93.874 - 80.421

Energy dissipated = 13.453 J

Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance between them is reduced to r0/4 ?What is the electrostatic force between and electron and a proton separated by 0.1 mm?

Answers

Answer:

The electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$

Explanation:

It is given that, charge $q_1$ is placed at a distance $r_o$ from charge $q_2$ . The force acting between charges is given by :

$F=(kq_1q_2)/(r_o^2)$

We need to find the force if the distance between them is reduced to $r_o/4$ . It is given by :

$F'=(kq_1q_2)/((r_o/4)^2)$

$F'=16* (kq_1q_2)/(r_o^2)$

$F'=16* F$

So, if the the distance between them is reduced to $r_o/4$ , the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or $10^(-4)\ m$ is :

$F=(kq_1q_2)/(r_o^2)$

$F=(9* 10^9* (1.6* 10^(-19))^2)/((10^(-4))^2)$

$F=2.30* 10^(-20)\ N$

So, the electrostatic force between and electron and a proton is $F=2.30* 10^(-20)\ N$ . Hence, this is the required solution.

Which one defines force?

Answers

Answer:

Explanation:

a push or a pull that occurs when an object interacts with another object or field.

pls mrk me brainliest

An isotropic point source emits light at wavelength 500 nm, at the rate of 185 W. A light detector is positioned 380 m from the source. What is the maximum rate ∂B/∂t at which the magnetic component of the light changes with time at the detector's location?

Answers

Answer:

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

Explanation:

Given that

An isotropic point source emits light at a wavelength $\lambda$ = 500 nm

Power = 185 W

Radius = 380 m

Let's first calculate the The intensity of the wave , which is = $(Power )/(Area)$

= $(Power)/(4 \pi r ^2)$

= $(185 \ W)/( 4 \pi (380)^2)$

= $1.0195*10^(-4) \ W/m^2$

Now;

The amplitude of the magnetic field is calculated afterwards by using poynting vector

i.e

$I = ((c)/(2 \mu_0 ))B_(max^2)$

$B_(max^2) = ((2 \mu_0 I)/( c))$

$B_(max^2) = ((2 *4 \pi *10^(-7)*1.0195*10^(-4))/( 3*10^8))$

$B_(max^2) = 8.5409*10^(-19)$

$B_(max) = \sqrt {8.5409*10^(-19)}$

$B_(max) = 9.242*10^(-10)$

The magnetic field wave equation can now be expressed as;

$B = B_(max) sin (kx - \omega t)$

Taking the differentiation

$(dB)/(dt)= - \omega B_(max) \ cos ( kx - \omega t)$

The maximum value ;

$(dB)/(dt) = \omega B _(max)$

where ;

$\omega = 2 \pi f\n\omega = (2 \pi c)/(\lambda)$

then

$(dB)/(dt) = (2 \pi c)/(\lambda) B _(max)$

$(dB)/(dt) = (2 \pi 3*10^8*9.242*10^(-10))/(500*10^(-9))$

$(dB)/(dt) = 3484751.917$

$(dB)/(dt) = 3.49 *10^(6) \ \ T/s$

The maximum rate $(∂B/∂t)$ at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

To find the maximum rate at which the magnetic component of the light changes with time at the detector's location, you can use the formula for the rate of change of magnetic field due to an electromagnetic wave. The formula is given by:

$∂B/∂t = (2π / λ) * E * c$

Where:

$∂B/∂t$ is the rate of change of the magnetic field.

λ is the wavelength of the light.

E is the electric field strength.

c is the speed of light in a vacuum, approximately $3 x 10^8 m/s.$

You have the wavelength (λ) as 500 nm, which is 500 x 10^-9 meters, and the electric field strength (E) can be calculated using the power (P) and the distance (r) from the source. The power emitted by the source is 185 W, and the distance from the source to the detector is 380 m.

First, calculate the electric field strength (E):

$E = sqrt(P / (2π * r^2))$

$E = sqrt(185 W / (2π * (380 m)^2))$

$E = sqrt(185 W / (2π * 144400 m^2))$

$E ≈ 6.325 x 10^-5 N/C$

Now, you can calculate the rate of change of the magnetic field:

$∂B/∂t = (2π / λ) * E * c$

$∂B/∂t = (2π / (500 x 10^-9 m)) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ (3.77 x 10^15 Hz) * (6.325 x 10^-5 N/C) * (3 x 10^8 m/s)$

$∂B/∂t ≈ 6.8 x 10^9 T/s$

So, the maximum rate at which the magnetic component of the light changes with time at the detector's location is approximately $6.8 x 10^9$ Tesla per second (T/s).

Learn more about magnetic component of the light here:

brainly.com/question/33261927

#SPJ6

insulator allows the electric current to pass through it true or false please anyone please please please

Answers

I say the answer is eating chicken. Eating chicken provides many nutrients such as, protein, diabetes, and somminilla.

You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You place the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces

Answers

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).

Other Questions

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower. The electric field amplitude of the signal at that point is 800 mV/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. The intensity of the radio signal at that point is closest to

Explain what happent to the pressure exerted by an object when the area over which it is exerted:a) increase b) decrease

Which best describes a reference frame?