I need a diagram for how a scrap heap magnet works

A force of 200N is being applied over an area measuring 0.75m^2

Sonbull [250]

Answer:

357KG

Explanation:

3 0

3 years ago

Unpolarized light passes through two polarizers. Find the fraction of light from the first polarizer that gets through the secon

meriva

Answer:

0.5

Explanation:

Data provided in the question:

The angle between their transmission axes, θ = 60°

Now,

We have the relation,

I₁ = I₀cos²θ

where,

I₁ is the intensity of the transmitted light

I₀ is the intensity of the incident light

on rearranging, we get

$\frac{I_1}{I_0}$ =cos²60°

or

$\frac{I_1}{I_0}$ =0.5

8 0

3 years ago

Read 2 more answers

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at

vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

$g=-9.8 m/s^2$ (acceleration of gravity)

The initial vertical velocity is

$u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s$

where the negative sign means the direction is downward.

The vertical position of the ball is given by

$y(t) = h + u_y t + \frac{1}{2}gt^2$

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

$0 = 4.50 -1.96t-4.9t^2$

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

$v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s$

where u = 3.05 m/s is the initial speed and $\theta$ the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

$d=v_x t$

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

$d=(2.34)(0.78)=1.83 m$

7 0

3 years ago

The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.

max2010maxim [7]

Answer:

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg , L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3

α = 1.193 rad / s²

ac = α *2 ⇒ ac = 1.193 rad / s² * 2

ac = 2.86 m / s²

4 0

3 years ago

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

3 years ago