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3 years ago

What is the speed of a 3 cm diameter roll if it has a velocity of 1220 yards/hour?

Mathematics

1 answer:

Alex Ar [27]3 years ago

5 0

Answer:

The speed of the roll is 1239.52 revolutions per minut (1239.52 rpm).

Step-by-step explanation:

In this problem we want to know the angular speed of the roll, knowing its tangencial speed (1220 yards/hour).

The relation between the angular speed and the tangencial speed is

$v=\omega*r=\omega*(D/2)$

$\omega=2v/D$

In this case we have a diameter of 3 cm and a velocity of 1220 yards/hour. So we can write

$\omega=2v/D=2*1220\frac{yards}{hour} *\frac{1}{3\,cm} *\frac{91.44cm}{1 yard} *\frac{1hour}{60min}= 1239.52 \frac{1}{min}=1239.52 rpm$

So the speed of the roll is 1239.52 revolutions per minut (1239.52 rpm).

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Read 2 more answers

The line 3x-8y=k cuts the x-axis and y- axis at the point A and B respectively. If the area of ∆AOB =12sq.units, find the value

Sladkaya [172]

Given:

The equation is:

$3x-8y=k$

It cuts the x-axis and y- axis at the point A and B respectively.

The area of ∆AOB =12 sq.units.

To find:

The value of <em>k</em>.

Solution:

We have,

$3x-8y=k$

Substituting $x=0$ to find the y-intercept.

$3(0)-8y=k$

$0-8y=k$

$y=\dfrac{k}{-8}$

$y=-\dfrac{k}{8}$

Substituting $y=0$ to find the x-intercept.

$3x-8(0)=k$

$3x-0=k$

$x=\dfrac{k}{3}$

Area of a triangle is:

$A=\dfrac{1}{2}\times base\times height$

The height of the ∆AOB is $OB=\dfrac{k}{8}$ because distance cannot be negative and the base of the ∆AOB is $OA=\dfrac{k}{3}$ . So, the area of the ∆AOB is: