What does periodic mean

A push on a textbook with 10 N moves it to a distance of 2 cm. How much force is need to move it to 4 cm?

uysha [10]

Answer:

20N

Explanation:

Ratio of N to cm-

10:2

so to make 2=4 times 2 so The ratio is now-

20:4

so to move 4 cm you need to push 20N.

8 0

3 years ago

A 215-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

Misha Larkins [42]

Answer:

303.9481875 N

Explanation:

t = Time taken = 2 seconds

F = Force

r = Radius = 1.5 m

I = Moment of Inertia

$\alpha$ = Angular Acceleration

Torque

$\tau=F\times r$

$\tau=I\times \alpha$

$\\\Rightarrow F\times r=I\times \alpha\\\Rightarrow F=\frac{I\times \alpha}{r}$

Angular velocity

$\omega=rev/s\times 2\pi\\\Rightarrow \omega=0.6\times 2\pi\\\Rightarrow \omega=3.76991\ rad/s$

Angular acceleration

$\alpha=\frac{\omega}{t}\\\Rightarrow \alpha=\frac{3.76991}{2}\\\Rightarrow \alpha=1.88495\ rad/s^2$

$I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}215\times 1.5^2\\\Rightarrow I=241.875\ kgm^2$

$F=\frac{I\times \alpha}{r}\\\Rightarrow F=\frac{241.875\times 1.88495}{1.5}\\\Rightarrow F=303.9481875\ N$

The magnitude of the force to stop the merry-go-round is 303.9481875 N

3 0

3 years ago

A pallet of bricks is to be suspended by attaching a rope to it and connecting the other end to a couple of heavy crates on the

SashulF [63]

Answer:

<h3>C.250 lb is the answer...</h3>

5 0

3 years ago

The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob

Crazy boy [7]

Answer:

104.3 cm or 179.7

Explanation:

First find time that it takes for the object to hit the ground

$\sqrt{(2H)/g} -> \sqrt{(2 x 179)/ 9.8} = 6.04s\\$ *

Then find xf of projectile $xf= 5.9(6.04) = 37.7\\\\$

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7 or 142+37.7

hope that helps

4 0

3 years ago

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 5

LekaFEV [45]

Answer:

$|\vec{F}_3| = 102.92 \ N$

$\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$ .

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$ .

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$ ,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$ .

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$ ,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$ ,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.

7 0

3 years ago