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3 years ago

How many mm are in a micro-m?

1 answer:

4 0

For this case we have that by definition, a micrometer is equivalent to a thousandth of a millimeter, that is, 0.001 millimeters.

Then, in other words, we have 0.001 millimeters in a micrometer.

Answer:

In a micrometer there are 0.001 millimeters.

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A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid

tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = $\frac{ r x ω^{2} }{g}$

where ω (angular velocity) = $\frac{2π}{time}$

= $\frac{2π }{5.5}$ = 1.14

Uk = $\frac{ 5.7 x 1.14^{2} }{9.8}$ = 0.76

6 0

3 years ago

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

VladimirAG [237]

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

<u>Explanation:</u>

Given data,

E= 3 ×10 ⁶ Δx=0.06/100

We have to find the minimum potential difference

E= -ΔV/Δx

ΔV=- E × Δx

ΔV =-3 ×10 ⁶ . 0.06/100

ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

6 0

3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road