An electric field is described by the vector e=yj -xi. what is the electric flux through a rectangular box in the x-z plane that is bounded by x between 0 and 1, and z between 0 and 1? (You need to identify dA to do this problem.)

Considering the Watt's law, if a current uses 31 amperes and has a voltage of 110 volts, it dissipates 3,410 Watts.

Current is the physical magnitude that expresses the amount of electricity that flows through a conductor in unit time and is measured in amps.

The voltage is the difference between the electrical charge that leaves the source and the one that finally reaches the end of the circuit. It is expressed in volts.

Watt's Law refers to the electrical power of an electronic component or device and is defined as the power consumed by the load is directly proportional to the voltage supplied and the current flowing through it. The unit of power is the Watt.

Knowing the voltage and current, this law is expressed as:

P = V×I

In this case, you know:

Replacing in Watt's Law:

P = 110 volts× 31 amperes

Solving:

P= 3,410 Watt

Finally, if a current uses 31 amperes and has a voltage of 110 volts, it dissipates 3,410 Watts.

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The radius of a helium atom 4.8 times bigger  then the radius of an alpha particle.

The structure of alpha particles, also known as alpha rays and alpha radiation, is similar to that of the helium-4 nucleus and has been made up of two protons as well as two neutrons bonded together.

For welding metals like aluminum, helium was utilized as an inert gas environment. It is also employed in rocket propulsion.

The radius of a helium atom 4.8 times bigger  then the radius of an alpha particle. More precisely than it's ever been, the diameter of the helium atom's nucleus, the alpha particle, had also been measured. Outcomes just point to a size of 1.6782 femtometers, 4.8 times more accurate than earlier readings.

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Answer:

Pressure will triple   (Boyle's Law)  

Explanation:

Assuming a constant temperature:  (compressing gases usually raises the temp significantly):

P1V1 = P2V2

P1V1 / V2   = P2      

   Now change V2  to 1/3 V2:

P1V1 / (1/3 V2 ) = P2 / (1/3 )

P1V1/ (1/3 v2 ) = 3 P2                <======= THE PRESSURE WILL TRIPLE

    This is Boyle's Law

So, the complete sentence is If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

When the mass of the sun is larger, Earth moves around the sun at a faster pace and When the mass of the sun is smaller, Earth moves around the sun at a slower pace.

When Earth is closer to the sun, its orbit becomes faster and When Earth is farther from the sun, its orbit becomes slower.

When Earth is closer to the sun, there will be a hotter climate. A little movement that takes one closer to the sun could lead to a huge impact, as the sun is very hot.

So, it can be concluded that If the mass of the sun is 2x, at least one planet will fall into the habitable zone if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully for the best conditions.

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Answer:

Tolerate

Explanation:

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time [tex]t[/tex], the horizontal position [tex]x[/tex] and vertical position [tex]y[/tex] of the ball are given respectively by

[tex]x = v_i \cos(\theta_i) t[/tex]

[tex]y = v_i \sin(\theta_i) t - \dfrac g2 t^2[/tex]

and the horizontal velocity [tex]v_x[/tex] and vertical velocity [tex]v_y[/tex] are

[tex]v_x = v_i \cos(\theta_i)[/tex]

[tex]v_y = v_i \sin(\theta_i) - gt[/tex]

The ball reaches its maximum height with [tex]v_y=0[/tex]. At this point, the ball has zero vertical velocity. This happens when

[tex]v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g[/tex]

which means

[tex]y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g[/tex]

At the same time, the ball will have traveled half its horizontal range, so

[tex]x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g[/tex]

Solve for [tex]v_i[/tex] and [tex]\theta_i[/tex] :

[tex]\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0[/tex]

Since [tex]0^\circ<\theta_i<90^\circ[/tex], we cannot have [tex]\sin(\theta_i)=0[/tex], so we're left with (e)

[tex]3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}[/tex]

Now,

[tex]\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}[/tex]

[tex]\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}[/tex]

so it follows that (d)

[tex]R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}[/tex]

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

[tex]v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}[/tex]

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

[tex]v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}[/tex]

Then with [tex]v_y=0[/tex], the ball's speed [tex]v[/tex] is

[tex]v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}[/tex]

Finally, in the work leading up to part (e), we showed the time to maximum height is

[tex]t = \dfrac{v_i \sin(\theta_i)}g[/tex]

but this is just half the total time the ball spends in the air. The total airtime is then

[tex]2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}[/tex]

and the ball is in the air over the interval (a)

[tex]\boxed{0 < t < 2\sqrt{\frac R{3g}}}[/tex]

Answer:

Multiple so you can test multiple hypothesis at once

Explanation:

because

Answer:

To test the hypothesis, we need to make an observation or perform an experiment associated with the prediction. For instance, in this case, we would plug the toaster into a different outlet and see if it toasts.

inspired by answer up there

sorry i couldn't add details because think its bad

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

Based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.

From the standpoint of  brain lateralization, differences do exist such as based on experience such as handedness or specific skills such as playing a guitar.

Note that Left-handers are said to have reduced or little lateralized brains, which tells us that the two halves of the brain are little different than as seen in the right-handers.

Therefore, I can say that based on my personal experiences, I believe that there is a difference between areas in precision of control because there are different part of the brain that controls the functions of the body.

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Answer:

Joes displacement is 54m

Answer:

1 / f = 1 / o + 1 / i = (i + o) / o * 1

f = o * i / (o + i) = 60 * 30 / (60 + 30) = 1800 / 90 = 20 cm

Both the object and image are in positive space for a mirror

Typing in the commands and instructions and pressing enter will put a new command in the command prompt​.

This refers to a words or phrase which causes the computer to execute certain tasks or functions.

Typing the exact instruction and entering it will ensure that the chosen tasks are found in the command prompt.

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The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.

Three equation of motion are:-

Where v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.

In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.

Applying equation 1 to find the initial speed of plane

v = u + at

0 = u + (-6.34 × 5400)   {v=0 as plane will stop after 5400 sec}

u =  6.34 × 5400

u = 34236 m/s

Initial velocity of plane is 34236 m/s

Applying equation 2 to find the displacement of plane in that time period

s = ut + (1/2)at²

s = ( 34236 × 5400 )  - ( (1/2) × 6.34 × 5400² )

s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )

s = 5400 × ( 34236 - 17118 )

s = 5400 × 17118 metres

s = 5.4 × 17118 Km

s = 92437.2 Km

Distance travelled by plane is 92437.2 Km

So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.

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