4025.05m +20.0m +0.050004m

Two people push on the same door from opposite sides as shown they will only see the door move when

rusak2 [61]

One of them overpowers the other

5 0

4 years ago

A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave

7nadin3 [17]

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

$I = \frac{1}{2} C{ {\varepsilon _0}E_m^2$

Here, $\varepsilon _0$ is the permittivity of the free space,

$E_m$ is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10 −12 C² /N⋅m² for ${\varepsilon _0}$

and 93.3 V/m for ${E_{\rm{m}$

$I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2$

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

$E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J$

The energy flowing through the given area is 4.55 J

5 0

4 years ago

A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor a

Zolol [24]

Answer:

v₀ = 6.64 m / s

Explanation:

This is a projectile throwing exercise

x = v₀ₓ t

y = y₀ + v_{oy} t - ½ g t²

In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m

the time to reach the basket is

t = x / v₀ₓ

we substitute

y- y₀ = $\frac{ v_o \ x \ sin \theta }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }$

y - y₀ = x tan θ - $\frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }$

we substitute the values

3 -1.8 = 3.0 tan 60 - $\frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}$

1.2 = 5.196 - 176.4 1 / v₀²

176.4 1 / v₀² = 3.996

v₀ = $\sqrt{ \frac{ 176.4}{3.996} }$

v₀ = 6.64 m / s

6 0

3 years ago

What will happen to the period of a ball-on-a-spring if the ball is replaced by a smaller one with half the mass

Murrr4er [49]

When the ball on a spring is replaced by a smaller ball with half the mass : The period decrease by 29%

<h3>Calculate the Period </h3>

We will apply the formula below to calculate period

T = $2\pi \sqrt{\frac{m}{k } }$

where :

T₁ / T₂ = $\sqrt{\frac{m_{1} }{m_{2} } }$ where m₂ = m₁ / 2

Therefore :

T₂ = $\frac{T_{1} }{\sqrt{2} }$ = 0.71 T

Hence we can conclude that the period will decrease by 29% ( i.e 100 - 71 ).

Learn more about Period calculation : brainly.com/question/10728818

4 0

2 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$

$Y=(24cm/tan(32.12))$

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0

3 years ago