This should be pretty easy i think.

Describe the components within the system involved in the collision between pilot and landing zone.

Answer:

it looks like dots and just black space on a large scale

Explanation:

on a large scale the universe especially our milky way looks small

hope this helps  

Answer:

in a furnished room the furniture scatter the sound waves and it doesn't reach our ears in the same manner it left our mouth.

She would be better off fitting in 15-20 minutes of exercise several times a week. This t is the BEST advice for Evonne.

Regular physical activity and exercise have many positive health effects that are difficult to deny. Everyone, regardless of age, sex, or physical ability, benefits from exercise.

Exercise can assist sustain weight loss or prevent excessive weight gain. Calorie burn occurs during physical exertion. You burn more calories when you engage in more vigorous exercise.

Numerous health issues and concerns, such as stroke, the metabolic syndrome, high blood pressure, and type 2 diabetes, can be prevented or managed with regular exercise. It can also help with cognitive improvement and reduce the danger of dying from any cause.

Exercise helps your circulatory system function more effectively and distributes oxygen and nutrients to your tissues. Additionally, you have greater energy to complete daily tasks as your heart and lung health improves.

Learn more about physical activity, refer the link:

https://brainly.com/question/24423889?

#SPJ2

Using visual inspection and knowledge of linear regression , the inconsistent input is the last data point (30, 12.0).

Therefore, the inconsistent data point is (30, 12.0)

Learn more : https://brainly.com/question/25651592

Initial kinetic energy KE = 1/2 m1v12 + 1/2 m2v22 = joules. The following calculation expects you to enter a final velocity for mass m1 and then it calculates the final velocity of the other mass required to conserve momentum and calculates the kinetic energy either gained or lost to make possible such a collision.

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 5 × 10 = 50

We have the final answer as

Hope this helps you

all of the choices.

Explanation:

....

Answer:

weight =m x g =2kg × 10m/s 2 = 20N

weight = m × gravity

=unit for weight is newtons

Mass=weight ÷ gravity

= unit for mass is kilograms

Gravity=weight ÷ mass

=unit for gravity is 10m/sec 2

Answer:

kicking ball and throwing stone

Answer:

- Pushing a shopping cart

- Kicking a soccer ball

Explanation:

    The second law states "It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it."

    So two examples would be pushing / kicking anything (shopping cart or soccer ball for example) and riding a bicycle (bike = mass and you are the force)

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

The cyclist will make it since the distance  traveled by the cyclist is greater than the length of the muddy stretch.

The given parameters;

The effective acceleration of the cyclist at the given coefficient of friction is calculated as follows;

[tex]a = \ \mu g\\\\a = 0.6 \times 9.8\\\\a = 5.88 \ m/s^2[/tex]

The distance traveled by the cyclist at the give speed and acceleration is calculated as;

[tex]v^2 = u^2 + 2as\\\\12^2 = 0 + 2(5.88)s \\\\ 11.76 s = 144\\\\ s = \frac{144}{11.76} \\\\a = 12.24 \ m[/tex]

Thus, we can conclude that the cyclist will make it since the distance  traveled by the cyclist is greater than the length of the muddy stretch.

Learn more here:https://brainly.com/question/14121363

Answer:

The thermosphere experiences a temperature increase primarily due to the absorption of gamma-rays and X-rays.

The characteristics of the atomic spectrum allow to find the result so that the measured lines help the astronomer is:

The emission spectrum is the emission of light due to the atomic transition in atoms, consequently the spectrum of each chemical element is unique.

The astronomer was able to identify the chemical elements that are in stars by their emission spectrum.

In the case presented there are two lines in the blue region, a line in the green region and a line in the red region. We know that the most abundant element in stars is hydrogen.

Hydrogen has a series of well-defined spectral lines, the series that is in the visible part of the spectrum is called the Balmer series and is composed of the lines in the table.

Color     wavelength (nm)

red           656.3

blue          486.1

violet        434.1

The second element in quantity in stars is helium, which has the spectrum given in the table and in the attached.

color   wavelength (nm)

Red          667.8

orange     587.6

green       501.6

Blue         492.2

Blue         447.1

The astronomer compares his measured spectrum with the spectra of these two gases and finds that the element he is measuring is helium.

In conclusion, using the characteristics of the atomic spectra, we can find the result to help the astronomer with the measured lines is:

Learn more here:  brainly.com/question/14819167

Answer:b

Explanation:

did it

Answer:

you are very kind Po lang

Answer:

current forms when electromagnetic waves strike a semiconductor, removing some of its electrons.

Explanation:

Answer:

current forms when electromagnetic waves strike a semiconductor, removing some of its electrons.

trust me i just did it

Explanation:

Answer: The Senate provides advice and consent on presidential nominations.

Answer:

b i think

Explanation:

hope this helps

Answer:  A B C or D which one

Explanation:

The charge flow if the energy transferred is 5.25 kJ and the potential difference is 15 volts is 350 Coulomb

Potential difference is the amount of work needed to move a unit charge from one point to another. It is given by:

V = E/Q

Where V is potential difference, E is energy and Q is charge. Given that:

V = 15 V, E = 5.25 kJ = 5.25 * 10³ J.

V = E/Q

15 = 5.25 * 10³ /Q

Q = 350 Coulomb

The charge flow if the energy transferred is 5.25 kJ and the potential difference is 15 volts is 350 Coulomb

Find out more at: https://brainly.com/question/3412043

Answer:

B. between carbon and oxygen

Explanation:

formaldehyde has a carbon atom that has single bonds with two hydrogen atoms on both sides, and it has a double bond with oxygen.

Answer:

The placement of the double bond is between carbon atom and oxygen atom

this same question was for me so I did and got it correct

please mark me as brainliest

Answer:

[tex]\huge\boxed{\sf t = 40\ seconds}[/tex]

Explanation:

Given data:

Acceleration = a = 0.5 m/s²

Initial Velocity = Vi = 0 m/s

Distance = S = 400 m

Required:

Time = t = ?

Formulas used:

2 formulas

Solution:

Using third equation of motion firstly

2as = Vf²-Vi²

2(0.5)(400) = Vf² - 0²

400 = Vf²

Taking sqrt

Vf = 20 m/s

Now, Using First equation of motion:

Vf = Vi + at

20 = 0 + 0.5t

0.5 t = 20

Divide both sides by 0.5

t = 40 seconds

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

e

Explanation:

It is possible to create a wave having some frequency on a rope and then gradually it will become faster. During this time wavelength becomes shorter. Here the wave 'e' is found to has highest frequency. The correct option is E.

The number of oscillations of a wave per unit time is defined as the frequency. It is usually measured in Hertz and the frequency is directly proportional to the pitch. Humans can hear sounds with frequencies ranging between 20-20000 Hz.

The sounds with frequencies over the human ears range are called the ultrasonic sound and sounds with frequencies less than the audible range are called infrasonic sound. The wavelength and frequency are inversely proportional to each other.

When wavelength increases, frequency decreases and vice versa. So here wave E has maximum number of vibrations.

Thus the correct option is E.

To know more about frequency, visit;

https://brainly.com/question/18125929

#SPJ2

Answer:

no picture attached

Explanation:

Answer:

A.)a=44.57m/s

B.)f=38,909.61

C.)f=666.4

Explanation:

A.) A=v/t 26.3/0.590=44.57

B.) 873kg x 44.57= 38,909.61

C.) 68kg x 9.8=666.4

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds, Acceleration is -2 m/s² (deceleration), and distance is 242 meters.

To solve this problem, it is given:

Initial velocity ([tex]\(v_i\)[/tex] ) = 44 m/s

Final velocity ([tex]\(v_f[/tex] ) = 22 m/s

Time ([tex]\(t\)[/tex] ) = 11 s

We have two tasks: finding the acceleration (a) and then using it to calculate the distance (d).

1. Calculate Acceleration (a):

[tex]\[a = \frac{v_f - v_i}{t} \\\\= \frac{22 \, \text{m/s} - 44 \, \text{m/s}}{11 \, \text{s}} \\\\= -2 \, \text{m/s}^2\][/tex]

The negative sign indicates deceleration (slowing down).

2. Calculate Distance (d):

Using the kinematic equation:

[tex]\[d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2\][/tex]

Plugging in the values:

[tex]\[d = (44 \, \text{m/s}) \cdot (11 \, \text{s}) + \frac{1}{2} \cdot (-2 \, \text{m/s}^2) \cdot (11 \, \text{s})^2\][/tex]

Calculating the first part:

[tex]\[d = 484 \, \text{m} - 121 \, \text{m} \\\\= 363 \, \text{m}\][/tex]

Calculating the second part:

[tex]\[d = 363 \, \text{m} + \frac{1}{2} \cdot (-2 \, \text{m/s}^2) \cdot (121 \, \text{s}^2) \\\\= 363 \, \text{m} - 121 \, \text{m} \\\\= 242 \, \text{m}\][/tex]

The total distance the race car moves during this time is 242 meters.

Thus, acceleration is -2 m/s² (deceleration), and distance is 242 meters.

For more details regarding acceleration, visit:

https://brainly.com/question/2303856

#SPJ3

Answer:

0.39 J/g°c

Explanation:

= heat / unit of mass × unit of temperature

986.75J/16.75g

= 58.9 J/g

∆T=175°c - 25°c = 150°c

986.75 / 150°c = 6.578

986.75 / 16.75g.150°c = 0.30 j/g°c

Answer:

The light waves are refracted as they enter and leave the prism. The shorter the wavelength of the light, the more it is refracted. As a result, red light is refracted the least and violet light is refracted the most - causing the coloured light to spread out to form a spectrum.

Explanation:

hope this helps

Answer: Neither, its actually violet aka purple. This is because it has the shortest wavelength. This is also why red light is refracted the least because it has the longest wavelength.

Would appreciate brainly <3

According to the Question:

[tex]\footnotesize\implies F \propto [m]^x [v]^y [r]^z[/tex]

[tex]\footnotesize\implies F =k [m]^x [v]^y [r]^z \:...(i)[/tex]

Now, substituting Dimensions of each physical quantity in equation (I):

[tex]\footnotesize\implies [M^1L^1T^{-2}] =[M^1L^0T^0]^x [M^0L^1T^{-1}]^y [M^0L^1T^0]^z[/tex]

[tex]\footnotesize\implies [M^1L^1T^{-2}] =[M]^x [L]^{y + z} [T]^{ - y}[/tex]

On comparing the powers of LHS and RHS:

[tex]\implies \bf x = 1[/tex]

[tex]\implies - y = - 2[/tex]

[tex] \implies \bf y = 2[/tex]

[tex]\implies y + z = 1[/tex]

[tex]\implies z = 1 - 2[/tex]

[tex]\implies \bf z = - 1[/tex]

[tex]\footnotesize\implies F = [m]^x [v]^y [r]^z[/tex]

[tex]\footnotesize\implies F = [m]^1 [v]^2 [r]^{ - 1}[/tex]

[tex]\implies\footnotesize \underline{ \boxed{ \bf \red{ F = \frac{m {v}^{2} }{r} }}}[/tex]

The age of the given rock is 700 million years.

The given parameters;

age of the rock ----------------------------- mass remaining

0 [tex](t_0)[/tex]  -------------------------------------------  20 g

1 half life ([tex]1 \ t_{1/2}[/tex] ) ------------------------------ 10 g

Age of the rock [tex]= t_{1/2} - t_0 = 700 \ - 0 = 700 \ million \ years[/tex]

Thus, the age of the given rock is 700 million years.

Learn more here:https://brainly.com/question/23774743

Answer:

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]

[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]