All categories

3 years ago

1mm is equal to .1 what

Physics

1 answer:

bulgar [2K]3 years ago

8 0

Answer:

When you have to do an English-Metric (SI) length conversion, and you already know the English units of length (miles, yards, feet, inches, etc.), all you need to remember is one simple relationship, and you can readily convert any length in the SI system, to the equivalent length in the other.

1 foot (ft) = 0.3048 meters (m)

BIn this case you need your answer in inches. You (hopefully) know there are 12 inches in a foot, so you just do the following:

1 inch (in) = 1/12 ft = 0.3048/12 m = 0.0254 m

You might be interested in

The rate at which the earth's surface loses heat and keeps earth temperature at which living things can survive

agasfer [191]

I believe it's insulation.

5 0

3 years ago

Black holes must have a mass three or more times the mass of _____.

tia_tia [17]

The answer is d. the sun

7 0

3 years ago

Read 2 more answers

The old rubber boot has two leaks. The top of the boot is 0.3 m higher than the leaks. What is the velocity of the water coming

Otrada [13]

Answer:

Leak 1 = 3.43 m/s

Leak 2 = 2.42 m/s

Explanation:

Given that the top of the boot is 0.3 m higher than the leaks.

Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2

From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.

Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2

Find the attached files for the solution and figure

7 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

Diagonal Launch

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$