A force of 15N is exerted on an area of 0.1 m squared. what is the pressure

In other words, it would take Deep Space 1 more than 81,000 years to travel the 4.24 light-years between Earth and Proxima Centauri at its top speed of 56,000 km/h. In relation to human history, that would be more than 2,700 generations.

Nearly 40 trillion kilometers, or 4.4 light-years, separate us from Alpha Centauri. The NASA-Germany Helios probes, the fastest spacecraft to date to be launched into orbit, flew at a speed of 250,000 kilometers per hour. The probes would need 18,000 years to travel at such pace to arrive at the sun's nearest neighbor. The calculations reveal that it is almost impossible to reach the nearest star in a human lifetime, even with the most futuristic technologies.

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Considering a scenario in which the Sun is going to explode.

In order to escape the explosion, we depart in a spacecraft with a speed of v = 0.8c.

The star Tau Ceti is 12 life years away.

At the midpoint of the journey, the Sun as well as Tau Cetik explode. at the same instant.

Imagine a hermit who is immobile with respect to both the Sun and Tau Ceti and resides on an asteroid that is midway between both. He observes the blast waves of both explosions as our spaceship passes him. This observer concludes that the two stars blew up simultaneously because he believes both stars to remain stationary.

In the frame of reference, the Sun and Tau Ceti explode at the same time and will remain stationary.

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Newton's Second Law of Motion states "The rate of change of momentum of a body is directly proportional to the applied unbalanced force in the direction of force".

Answer:

the circles are called Sun spots.  they are colder than the surrounding area.

Explanation:

Answer:

The circled areas are called__Sunspots___. They are___colder____ the surrounding area.

Explanation:

i just took the test it is correst plz give me a thanks or a brainiest and ill give 1 back :D

He should repeat the test with three additional batteries of varying voltages to demonstrate the reproducibility of results. he should use only one battery but do several trials, using different circuit designs to ensure he has appropriate test groups - this can improve the student's experiment.

Reproducibility, otherwise called replicability and repeatability, is a significant rule supporting the logical technique. For the discoveries of a review to be reproducible implies that results got by an examination or an observational review or in a factual investigation of an informational index. Albeit numerous organic researchers instinctively accept that the reproducibility of a trial implies that it tends to be repeated, Drummond makes a differentiation between these two terms. Informational indexes are huge and calculations are more refined. This requests reproducibility to decrease the mistake and predisposition when people are added to the course of information investigation.

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They are maintaining their body temperature by cooling down, like dogs do, as they don't have enough sweat glands to keep them cool without panting.

Brown bears maintain body temperature by cooling as a result of panting. So, the correct option is D.

Normal human body-temperature is defined as the typical temperature range found in humans which is commonly referred to as 36.5–37.5 °C. The human body temperature varies. Temperature is known as the degree of hotness or coldness of an area or object.

When we are very hot, the blood vessels of the skin dilate to carry the excess heat to the surface of the skin. Brown bear panting ensures that cool air circulates over the moist surfaces in the mouth which helps maintain body temperature by cooling the body.

Therefore, the correct option is D.

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Answer:

It takes 0.416 seconds to hit the ground.

Explanation:

speed = distance/ time

12 = 5/ x (multiply both with x)

12x = 5 (÷ 12)

x = 0.416 s

Answer: True

Explanation:

Answer:

false

Explanation:

hope this help (brainlist plz)

Answer:

[tex]v_x=34 m/s[/tex]

[tex]v_y=53.9\ m/s[/tex]

Explanation:

Horizontal Launch

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

[tex]v_y=g.t[/tex]

The horizontal component of the velocity is always the same:

[tex]v_x=34 m/s[/tex]

The vertical component at t=5.5 s is:

[tex]v_y=9.8*5.5=53.9[/tex]

[tex]v_y=53.9\ m/s[/tex]

Answer:

U^U

Explanation:

Two-dimensional vectors have two components: an x vector and a y vector. Each of these vector components is a vector in the direction of one axis. The sum of the components of vectors is the original vector. Three-dimensional vectors have a z component as well.

Answer:

B. Applied

Explanation:

Applied means that somethings is pushing directly into it like a person to a box. and well the reasons why the others are not contact force are because gravity its cause to an attraction to the planet and no matter where you are it will still attract you, electric is not because it can flow through air, magnetic it affects to an invisible field that even if its not torching the marital you can still feel the magnetic push or pull

So applied

Adding the rise time and fall time then the package to reach the ground in 3.647 second.

Time can be defined in the sence as,that it is a continued sequence of existence and events that occurs in an apparently irreversible succession from the past, through the present, into the future.

We are given that ,

The packege 's trajectory into a rise to maximum height and fall from there to the ground .Rise to maximum height begins at vertical position is  given as,

Vertical position of the helicopter = H₀ = 20 m,

Initial velocityof helicopter = u = 14 m/s upward

Covered time of the helicopter =t₁ = 0,

Acceleration under gravity of helicoper=(- a) = -9.81 m/s²

Rise ends when instantaneous velocity v = 0,

Therefore, from the equation of motion for velocity as a function of time is  written as ,

v = a + at

0 = 14 m/s - (9.81 )m/s² t₁

t₁ = 1.427sec

Thus , the velocity changes linearly over time t₁ sec from u = 0 m/s then the averaging velocity can be given by w i.e.

w = u /2 = (14m/s)/2

w = 7 m/s

Therefore, maximum height of packege is given as,

H₁ = H₀ + w × t₁

H₁ =20m + 9.81m/s (1.427)s

H₁ = 33. 998 m

Therefore, when the packege fall from maximum height begins at ,

Time= t₁ = 0,

Position= h = H₁

Velocity= u = 0, and

Acceleration due to gravity downward=(-a) = -9.81 m/s²

Thus the  equation of motion for position (s) as a function of initial position (h) is can be written as,

s = h + ut₁ + (1/2) at₂²

0 = H₁ + 0 + (1/2) (-9.81)t₂²

t₂² = 4. 890 s

t₂ = 2.21 s

Hence , on adding the rise time t₁ and fall time t₂

Total time T = t₁ + t₂

T = 1.427 s + 2.21s = 3.647 s

Thus, then the package to reach the ground in 3.647 second.

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The aircraft takes 25s to become airborne and will cover a distance of 1.0417km.

Distance, s = [tex]\frac{300km}{1 h}[/tex]

                s = [tex]\frac{300 }{1}[/tex] × [tex]\frac{1000m}{3600s}[/tex]

                s = [tex]\frac{250}{3}[/tex] m/s

Velocity is the time derivative of the displacement, [tex]v(t)=\frac{d}{dt}(\frac{5}{3}t^{2})[/tex]

                                                                                      [tex]v(t)=\frac{10}{3}t^{2}[/tex]

The time required to reach [tex]v=\frac{500}{3}[/tex] m/s,

                                              [tex]\frac{10}{3}t=\frac{250}{3}[/tex]

                                              [tex]t=25s[/tex]

Now, during this time, the aircraft travels a distance, [tex]D= \frac{5}{3} (25)^{2}[/tex]

                                                                                       [tex]D= 1041.667m[/tex]

                                                                                      [tex]D= 1.0417km[/tex]

Therefore, the aircraft takes 25s to become airborne and will cover a distance of 1.0417km.

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Answer:

ewwwwwwwwwwww

get it out!

Answer:

C.

Explanation:

kilograms and m/s^2

The nfl caliber wide receiver that can reach a top speed of 10.0 m/s in only 2.4 s has an acceleration during this time of: 4.16 m/s²

The formula and procedure we will use to solve this exercise is:

a = (vf - vi) /t

Where:

Information about the problem:

Applying the acceleration formula we have:

a = (vf - vi) /t

a = (10.0 m/s - 0 m/s) /2.4 s

a = (10.0 m/s) /2.4 s

a = 4.16 m/s²

It is a physical quantity that indicates the variation of velocity as a function of time, it is expressed in units of distance per time squared e.g.: m/sec2 ; km/h2

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Answer:

The net force acting on the mass is 16.523 N

Explanation:

The given mass of the object, m = 1.3 kg

The length of the inclined plane over which the mass slides = 8.7 × 10⁻¹ m

The time it takes the mass to slide = 0.37 s

Therefore, we apply the following kinematic equation of motion;

s = u·t + 1/2·a·t²

Where;

u = The initial velocity of the mass = 0 m/s

t = The time taken for the motion = 0.37 s

a = The acceleration

s = The distance moved in the motion = 8.7 × 10⁻¹ m = 0.87 m

Plugging in the values gives;

0.87 = 0 × t + 1/2 × a × 0.37²

∴ 0.87 = 1/2 × a × 0.37²

a = 0.87/(1/2 × 0.37²) ≈ 12.71 m/s²

The net force acting on the mass is F = Mass, m × Acceleration, a force = m × a

Therefore, the net force acting on the mass is F = 1.3 × 12.71 = 16.523

The net force acting on the mass, F = 16.523 N.

The net force acting on the mass along the incline is 16.523 N

From the information given:

According to the second equation of motion;

[tex]\mathbf{S = ut + \dfrac{1}{2}at^2}[/tex]

the initial velocity u = 0 m/s since the mass starts from the rest

∴

[tex]\mathbf{S =0(t) + \dfrac{1}{2}at^2}[/tex]

[tex]\mathbf{S = \dfrac{1}{2}at^2}[/tex]

2S = at²

[tex]\mathbf{a = \dfrac{2S}{t^2}}[/tex]

[tex]\mathbf{a = \dfrac{2\times 0.87 \ m}{0.37^2}}[/tex]

acceleration (a) = 12.71 m/s²

The net force now acting on the mass along the incline can now be estimated as:

[tex]\mathbf{F_{net} = ma}[/tex]

[tex]\mathbf{F_{net} = 1.3 \ kg \times 12.71 \ m/s^2}[/tex]

[tex]\mathbf{F_{net} = 16.523 \ N}[/tex]

Therefore, we can conclude that the net force acting on the mass along the incline is 16.523 N

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Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

m is the mass

f is the force

From the question we have

[tex]a = \frac{56}{7} \\ [/tex]

We have the final answer as

Hope this helps you

The force is attractive.

The force between two parallel currents I₁ and I₂ that are spaced apart by r is measured as the ratio F/L. If the currents are flowing in the same direction, the force is attracting; if not, it is repulsive.

Each parallel wire carries 3.00 A of current in the same direction and is spaced apart by 6.00 cm.

Think of these two wires as being placed side by side on a table, with currents flowing in opposite directions toward you—wire 1 to the left and wire 2 to the right. The magnetic field produced by wire 1 at the location of wire 2 is directed vertically upward, according to the right-hand rule that links current to field direction. The force experienced by wire 2 as a result of its current flowing through this field is therefore determined by the right-hand rule, which links current and field to force, to be to the left, back toward wire 1. The force that one wire applies to another is therefore an attractive force.

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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

[tex]Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}[/tex]

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

[tex]E=\dfrac{Q}{4\pi\epsilon_0 r^2}[/tex]

Substitute numerical values:

[tex]E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}[/tex]

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

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The three different paper airplane designs were created by Ricardo. In an effort to determine which design flies the farthest, Ricardo intends to fly each paper airplane five times. Following each flight, Ricardo should measure the distance flown by the aircraft and note it.

Two of the most frequent techniques used in agricultural surveying involve measuring angles and distances with basic tools. even when the tools are basic. With diligence and practice, a great degree of precision can be attained.

The (1) pace is the main technique for gauging distance. 2) The odometer. (3) chaining or taping. four stages. optical range finder, number 5. and (6) techniques for electronic distance measurement (EDM).

It will be beneficial to review the standard units of distance first (displacement). The following are the standard English units:

12 inches make up one foot (in)

Yd (yard) = 3 feet (ft)

1 rod is equal to 16.5 feet and 5.5 yards (yd)

A mile is equal to 5280 feet, 1760 yards, and 320 rods.

When converting between different units of measurement, the units cancellation approach comes in quite handy.

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The ranking of the measurement from the largest to the smallest is 11.6 mg > 0.000006 kg > 1021 µg > 0.31 mg.

Ranking of measurement is a classification that describes the nature of information within the values assigned to variables.

This is usually done based on the size of each item in the list of measurement.

11.6 mg = 0.0116 g

1021 µg = 0.001021 g

0.000006 kg = 0.006 g

0.31 mg = 0.00031 g

Thus, the ranking of the measurement from the largest to the smallest is 11.6 mg > 0.000006 kg > 1021 µg > 0.31 mg.

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Answer:

A horse pulls a cart, a person walks on the ground

Explanation:

Answer:

Know it's late but I couldn't find a good answer when I looked for it so i did it myself.

My answer:

One of the real world examples mentioned in the article was that of a rocket blasting off into space. It was explained that it's an example of Newton's third law because the rocket’s engines push on those gases (action force), and the gases push back (reaction force). This reaction shows an excellent example of Newton's third law.

Sample Response:

When gases escape downward out the tail of a rocket, they push upward on the rocket. When layers of the Earth collide, they force the Earth upward, forming mountains.

Explanation:

hope this helps <3

Answer:

-49.49

Explanation:

If there’s no air resistance then the only thing acting on the object is the force of gravity which is mass times gravity. And to get mass all we have to do is divide 50.5/10 then times that by -9.8 since that’s the force of gravity.

Answer: The answer would be 50.5 N.

Explanation: It’s because there’s no air resistance.

The amplitude of the resultant wave in terms of the common amplitude [tex]y_m[/tex] of the two combining waves which are out of phase by [tex]\frac{\pi}{2} \ \text{rad}[/tex] is [tex]\sqrt{2} y_m[/tex].

The amplitude of two waves interfering with each other depends on the principle of superposition.

Since, the waves are out of phase by [tex]\frac{\pi}{2} \ \text{rad}[/tex] so we can write the waves to be:

[tex]y_1=Asin(kx-\omega t)[/tex]

[tex]y_2=Asin(kx-\omega t+\frac{\pi}{2})\\ \therefore y_2=Acos(kx-\omega t)[/tex]

According to the principle of superposition, the resultant wave is written as:

[tex]y=y_1+y_2[/tex]

[tex]\Rightarrow y=A[sin(kx-\omega t)+cos(kx-\omega t)]\\\Rightarrow y=\sqrt{2}A[\frac{1}{\sqrt{2}}sin(kx-\omega t)+\frac{1}{\sqrt{2}}cos(kx-\omega t)]\\y=\sqrt{2}Asin(kx-\omega t+ \frac{\pi}{4})[/tex]

So from the above relation, we have the amplitude to be:

New Amplitude = [tex]\sqrt{2}A[/tex]

Hence, the amplitude of the combined waves is [tex]\sqrt{2}A[/tex].

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Answer:

  0.6 seconds

Explanation:

The time to fall from height h is ...

  t = √(2h/g)

  t = √(2(1.86 m)/(9.8 m/s^2)) ≈ √0.3796 s ≈ 0.616 s

It would take about 0.6 seconds for the projectile to hit the ground.

Answer:

0.6 seconds

is ur answer :)

✌❤

Answer: c

Explanation:

Current is the flowing charges that goes around p.d increases it.

Answer:

C

Explanation:

Time taken to reach height is 4 seconds.

Given that;

Velocity of cannon ball = 80 m/s

Height taken = 320 meter

Find:

Time taken to reach height

Computation:

S = ut + (1/2)gt²

320 = (80)(t) + (1/2)(10)(t²)

320 = 80t + 5t²

t² + 16t + 64

So,

t = 4

Time taken = 4 seconds

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