All categories

3 years ago

What torque will increase angular velocity of a solid cylinder of mass 16 kg and diameter 1 m from zero to 120 rpm in 8 s?

Physics

1 answer:

const2013 [10]3 years ago

7 0

This is what I found online! Hope it helps:)

You might be interested in

Why does rubbing a balloons on wool or hair make it attract other objects

KonstantinChe [14]

Answer:

its because static electricity my guy

Explanation:

5 0

4 years ago

Read 2 more answers

A secant and a tangent meet at a 90Ă‚Â° angle outside the circle. What must be the difference between the measures of the interc

kozerog [31]

The difference between the measures of the intercepted arcs of the given circle is 45°.

A circle is defined as the set of points in a plane equidistant to each, such that it forms a closed two-dimensional figure, which is known as a circle.

A line intersecting a circle at a minimum of two distinct points is known as a secant and the line touching the circle at only one point, is known as the tangent of a circle.
The intercepted arc is the section of the circumference of a circle such that it is either encased by two chords or a line segment, meeting at a single point.

We are given that the angle subtended by secant and tangent, outside the circle is,

$\theta =90^{\circ}$

And angle measured inside the circle is,

$\phi = \theta /2\\\\\phi = 90/2\\\\\phi = 45{^\circ}$

So, the difference between the measures of the intercepted arcs is,

$\theta' = \theta - \phi\\\\\theta' = 90-45\\\\\theta' =45^{\circ}$

Thus, we can conclude that the difference between the measures of the intercepted arcs of the given circle is 45°.

learn more about the tangent here:

brainly.com/question/14022348

4 0

3 years ago

Ilya and Anya each can run at a speed of 7.30 mph and walk at a speed of 4.00 mph. They set off together on a route of length 5.

katrin [286]

Answer:

a)0.9675 h

b)5.168 mph

c)0.885 h

d)5.65 mph

Explanation:

The distance travelled by a moving object can be calculated as the velocity multiplied by the time, for this problem we can divide the problem in two stages, one where the are walking and another where they run,

Considering the total time for Anya is $t_{a}$ :

She first walks for half the distance (2.5 miles) so:

$2.5 miles=V_{w}*t_{a1}$

Solving for $t_{a1}$ :

$t_{a1}=\frac{2.5miles}{V_{w}}=\frac{2.5miles}{4 mph}=0.625 h$

Then she runs the other half of the distance:

$2.5 miles=V_{r}*t_{a2}$

$t_{a2}=\frac{2.5miles}{V_{r}}=\frac{2.5miles}{7.3 mph}=0.3425 h$

Adding this two durations we have the total time it takes for Anya to cover the distance:

$t_{a}=0.3425 h+0.625 h=0.9675 h$

The average velocity can be calculated as the total distance divided by the total time:

$V_{a-av} =\frac{5 miles}{0.9675 h} =5.168 mph$

Considering the total time for Ilya is $t_{I}$ :

She first walks for half the time so:

$d_{I1}=V_{w}*t_{I}/2$

Where $V_{w}$ is the velocity when walking (same for both people). On the second half she runs:

$d_{I2}=V_{r}*t_{I}/2$

This two distances must sum the total distance 5 miles, so:

$d_{I2}+d_{I2}=5 miles$

replacing the expressions above:

$V_{w}*t_{I}/2+V_{r}*t_{I}/2=(V_{w}+V_{r})*t_{I}/2=5 miles$

$t_{I}=\frac{5miles*2}{V_{w}+V_{r}}=\frac{10 miles}{4mph+7.3mph}=0.885 h$

so it takes Ilya 0.885 hours to to cover the distance

The average velocity can be calculated as the total distance divided by the total time:

$V_{I-av} =\frac{5 miles}{0.885 h} =5.65 mph$

3 0

3 years ago

Stars appear as small points of light in the sky because

Nonamiya [84]

Because they are so far away that the distance has to be measured in light years. Each light year is around 10 trillion kilometers, which is a very large distance.

8 0

3 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

$\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m$

This means that the relation between the wavelength and the length of the string is

$3\lambda/2 = L$

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

$y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)$

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

$v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)$

$a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0$

For this equation to be equal to zero, sin(59.94t) = 0. So,

$59.94t = \pi\\t = \pi/59.94 = 0.0524~s$

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

$v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s$

4 0

4 years ago