Using your Periodic Table, which of the elements below is most likely to be a solid at room temperature?

Mike recently purchased an optical telescope. Identify the part of the electromagnetic spectrum that is closest to the frequency

Zepler [3.9K]

The elctromagnetic spectrum ranges from the radiowaves to the gamma rays. The whole spectrum is shown in the attached picture. But the optical telescope can only see the visible region. So, it only covers from the 400 nm to 700 nm frequency. It follows the ROYGBIV colors, where red has the highest frequency and violet has the lowest frequency.

8 0

3 years ago

Read 2 more answers

Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or

slamgirl [31]

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

To learn more about feedback loop, visit:

brainly.com/question/11312580

#SPJ4

5 0

2 years ago

A) A spaceship passes you at a speed of 0.800c. You measure its length to be 31.2 m .How long would it be when at rest?

rosijanka [135]

Answer:

a

$l_o =52 \ m$

b

$l = 37.13 \ LY$

Explanation:

From the question we are told that

The speed of the spaceship is $v = 0.800c$

Here c is the speed of light with value $c = 3.0*10^{8} \ m/s$

The length is $l = 31.2 \ m$

The distance of the star for earth is $d = 145 \ light \ years$

The speed is $v_s = 2.90 *10^{8}$

Generally the from the length contraction equation we have that

$l = l_o \sqrt{1 -[\frac{v}{c } ]}$

Now the when at rest the length is $l_o$

So

$l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }$

$l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }$

$l_o=52 \ m$

Considering b

Applying above equation

$l =l_o \sqrt{1 - [\frac{v}{c } ]}$

Here $l_o =145 \ LY(light \ years )$

So

$l=145 * \sqrt{1 - \frac{v_s^2}{c^2 } }$

$l =145 * \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }$

$l = 37.13 \ LY$

4 0

2 years ago

You are in a car moving forward at 12 m/s. You throw a ball in the direction the car is moving. From your point of view, what do

4vir4ik [10]

Answer: it goes the same speed as the car

Explanation:

3 0

2 years ago

Scenario

Anvisha [2.4K]

Answer:

1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

0 = y₀ + 0 - ½ g t2

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 2650/9.8)

t = 23.26 s

this is the horizontal scrolling time

x = v₀ t

x = 366.67 23.26

x = 8527 m

the speed at the point of arrival is

v_y = v_{oy} - g t = 0 - gt

v_y = - 9.8 23.26

v_y = -227.95 m / s

Module and angle form

v = $\sqrt{v_x^2 + v_y^2}$

v = √(366.67² + 227.95²)

v = 431.75 m / s

θ = tan⁻¹ (v_y / vₓ)

θ = tan⁻¹ (227.95 / 366.67)

θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

cos θ = v₀ₓ / v₀

sin θ = v_{oy} / v₀

v₀ₓ = v₀ cos θ

v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + vo sin θ t - ½ g t²

0 = -50 + vo sin 50 t - ½ 9.8 t²

x = x₀ + v₀ₓ t

0 = x₀ + vo cos θ t

0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

50 = 0,766 vo t – 4,9 t²

400 = 0,643 vo t

resolved

50 = 0,766 ( $\frac{400}{0.643 \ t}$ ) t – 4,9 t²

50 = 476,52 t – 4,9 t²

t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

t = [97.25 ± $\sqrt{97.25^2 - 4 \ 10.2}$ ] / 2

t = 97.25 ±97.04] 2

t₁ = 97.145 s

t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

t = 97.145 s

let's find the initial velocity

x = x₀ + v₀ cos 50 t

0 = -400 + v₀ cos 50 97.145

v₀ = 400 / 62.44

v₀ = 6.4 m / s

5 0

2 years ago