I’m not sure what this is… help please

Answer:

Identify the name of the individual, group or institution that created the website. A reliable website should clearly state the name of its creator. Generally, websites created by government institutions (.gov) and educational institutions (.edu) are considered more reliable.

Hi there!

We can use impulse for this situation:

I = Δp = mΔv

Impulse = Force × time, so:

I = 63.9(24) = 1533.6 Ns

Find force by dividing by time:

I/t = 1533.6/1.2 = 1278 N

Carbon -14 is a dating method that is applicable to only samples of organic origin which are less than 50,000 years old.

When living things contain carbon- 14. Carbon - 14 is a radioactive isotope of carbon that decays by beta emission. As long as a plant or animal is alive, there is a balance between its intake and output of carbon -14.

However, when the plant or animal dies, this carbon - 14 begins to decay. The half life of carbon - 14 is 5700 years and it can be used to measure the age of materials or organic origin that are less than 50,000 years old.

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Answer:

the change of speed!

Explanation:

if light speeds up or slow down more it bend more

Answer:

refers to the bending of light

the fact or phenomenon of light, radio waves, etc. being deflected in passing obliquely through the interface between one medium and another or through a medium of varying density.

Answer:

Wavelength and speed both depend on the material of the medium.

Explanation:

Let [tex]f[/tex] denote the frequency of a wave. Let [tex]v[/tex] denote the speed of that wave in a given medium. Let [tex]\lambda[/tex] denote the wavelength of that wave in this medium.

The frequency [tex]f[/tex] of a wave stays the same regardless of the material of the media. However, the speed [tex]v[/tex] of a wave indeed depends on the material of the medium (density of the medium, etc.)

[tex]\lambda = v / f[/tex] and [tex]f[/tex] stays the same regardless of the medium. Thus, the wavelength [tex]\lambda[/tex] of this wave in a given medium would be proportional to the speed [tex]v[/tex] in that medium.

For example, the speed of sound in materials with a larger density (e.g., water) is generally larger than that in materials with lower density (e.g., air.)

When a sound wave travels from air into water, the frequency [tex]f[/tex] of this wave would stay the same. However, since the speed [tex]v[/tex] of the wave would increase, the wavelength [tex]\lambda[/tex] of this wave would also increase.

Answer:Gamma-Ray Telescopes

Explanation:

Gamma-Ray Telescopes are used for detecting the gamma-ray part of the Electromagnetic Spectrum. These can detect very high-energy gamma-rays from radio galaxies, gamma-ray bursts from stars, supernovae events, and Quasars.

Radio Waves Telescope

Explanation:

The main difference is frequency, sound waves are in Kilohertz , water waves are of course much slower. There are other differences , sound is longitudinal compression, but if you track motion of a particle in a water wave it moves in elliptic orbit as the wave passes.

The forces that push upward on an indoor skydiver are lift force and air resistance.

The forces that act on a skydiver moving downwards includes, gravity due to his weight, air resistance and lift force.

The downward forces on the indoor skydiver include the following;

The upward force on the indoor skydiver include the following;

Thus, we can conclude that the forces that push upward on an indoor skydiver are lift force and air resistance.

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The ratio of the electric force to the gravitational force between the two electrons is [tex]4.2 \times 10^{42}[/tex].

The given parameters;

The electrostatic force between the two electrons is calculated as follows;

[tex]F_c = \frac{kq^2}{r^2} \\\\[/tex]

The gravitational force between the two electrons is calculated as follows;

[tex]F_g = \frac{Gm_e^2}{r^2}[/tex]

The ratio of the electric force to the gravitational force between the two electrons is calculated as follows;

[tex]\frac{F_c}{F_g} = \frac{kq^2}{r^2} \times \frac{1}{\frac{Gm^2}{r^2} } \\\\\frac{F_c}{F_g} = \frac{kq^2}{r^2} \times \frac{r^2}{Gm^2} \\\\\frac{F_c}{F_g} = \frac{kq^2}{Gm^2}\\\\[/tex]

[tex]\frac{F_c}{F_g} = \frac{(8.99 \times 10^9) \times (1.6 \times 10^{-19}))^2}{(6.67 \times 10^{-11} ) \times (9.11 \times 10^{-31})^2} \\\\\frac{F_c}{F_g} = 4.2 \times 10^{42}[/tex]

Thus, the ratio of the electric force to the gravitational force between the two electrons is [tex]4.2 \times 10^{42}[/tex].

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The Correct statement is ~

Acceleration is a vector quantity.

Answer:

Acceleration is a vector quantity.

Answer:

the ratio of the number of turns in the primary and secondary coils.

Explanation:

hope i'm not too late!!

Answer:

chemical symbol

Explanation:

speed- the rate something moves or operates

Can I have brainlist please

The Speed are distance covered by the object in unit time.The S.I unot of speed is metre per second written as m/sec.It is a scalar quantity.

The speed is measured in metres per second ( ms- ¹ )

Where denoted by s

The distance travelled is measured in metres (m) and denoted by d

t is the amount of time spent and measured in seconds s

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

The net force acting on the mass along the incline is equal to 16 Newton.

Given the following data:

To determine the net force acting on the mass along the incline:

First of all, we would find the acceleration of the mass by using the second equation of motion:

[tex]S = ut+ \frac{1}{2} at^2\\\\4 = 0(1) + \frac{1}{2} a(1)^2\\\\4=\frac{a}{2}\\\\a=2\times 4[/tex]

Acceleration, a = 8 [tex]m/s^2[/tex]

For the net force, we would apply Newton's Second Law of Motion:

[tex]F_{net} = mass \times acceleration\\\\F_{net} =2 \times 8[/tex]

Net force = 16 Newton.

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Answer:

Very simple. The charge Q= 20 C. the formula for Q= I* t. now I= Q/t = 20/1s = 20 Amps.

Explanation:

The current given that the charge is 10C and the time is 4 seconds is gotten as Current = 2.5 Amps

Given Data

Charge = 10 Columbs

Time = 4 seconds

Current = ?

Electrical current is defined by how much electric charge has been transferred per second, giving the following relationship:

Charge = Current*Time

Rearranging this we get:

Current = Charge/Time

The symbol for charge is Q , it is measured in coulombs (C).

The symbol for current is I , it is measured in amperes (A).

The symbol for time is T, it is measured in seconds (s).

Substituting our given Data into the expression we have

Current = 10/4

Current = 2.5 Amps

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Answer: molecular solid

Explanation:

Answer:

it traveled 180km

Explanation:

Answer: gravity

Explanation:

Answer:

Seriously Your Question Is Kinda Tremendous

Explanation:

I'll Head Is Boiling From Question That Keeps Popping Up My Head, Someone Might Give You A Positive Hand

Answer:

Uranus

Explanation:

brainliest please?

You can't. Velocity and acceleration measure two different things, so their units are incompatible. It's like asking, "How many meters does this book weigh?"

Maybe you mean "find" acceleration using given velocities, or a velocity function?

Answer:

Explanation:

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

Answer:

b i think

Explanation:

hope this helps

Answer:  A B C or D which one

Explanation:

Hi there!

We can begin by calculating the acceleration using the kinematic equation:

d = vit + 1/2at²

Since the rocket starts from rest, we can rewrite this as:

d = 1/2at²

Plug in given values:

4.2 = 1/2a(0.137²) ⇒ a = 447.55 m/s²

Now, we can use Newton's Second Law to find the appropriate force:

∑F = m · a

∑F = 50,000 · 447.55 = 22,377,500 N

The beginning of the formation of galaxies and clusters of galaxies are the slight fluctuations seen in maps of the cosmic background radiation. So, the correct option is C.

The cosmic background radiation is defined as the microwave radiation that fills all of space which is a remnant that provides an important source of data on the early universe. With a standard optical telescope, the background space between stars and galaxies is almost completely dark.

It is also called the Cosmic Microwave Background (CMB) or electromagnetic radiation filling the universe which is a residual effect of the Big Bang that happened 13.8 billion years ago. The beginning of the formation of galaxies and clusters of galaxies are the slight fluctuations seen in maps of the cosmic background radiation.

Therefore, the correct option is C.

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Your question is incomplete, most probably the complete question is:

What are the slight fluctuations seen in maps of the cosmic background radiation?

a. uncertainties in the map

b. variations in the instrument's sensitivity

c. the beginning of the formation of galaxies and clusters of galaxies

d. dark matter

e. none of the above