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3 years ago

Please help me tbnmk,mnbnm

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

C perfect distance

Explanation:

This is untrue, life can exist from different locations from the sun such as titan or europa

kiruha [24]3 years ago

8 0

Answer:

Explanation:

Its C, because all the other planets have A, B, and D. But since earth is just far enough from the sun, it can support life. I think its called liek the perfect area section. Also, mars is too far, venus is the hottest planet, and mercury doesn't even have an atmosphere. im not going to go farther from the astroid belt.

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A 22-turn circular coil of wire has diameter 1.02 m. It is placed with its axis along the direction of the Earth's magnetic fiel

ollegr [7]

Answer:

ξ = 0.00845020162 V or 8.4 mV

Explanation:

Magnetic flux measures the total magnetic field that passes through a known area. Magnetic flux describe the effect of magnetic field in a given area. Mathematically,

magnetic flux (Ф) = BA cos ∅

where

A = test area

B = magnetic field

before the flip

Ф = Bπr²N

N = number of turn

magnitude of induced emf = N |ΔФ/Δt|

ξ = 2Nπr²B/dt

ξ = 2 × 22 × π × (1.02/2)² × 0.000047/0.2

ξ = 44 × π × 0.51² × 0.000047/0.2

ξ = 44 × π × 0.2601 × 0.000047/0.2

ξ = 0.0005378868 × 3.142/0.2

ξ = 0.00169004032/0.2

ξ = 0.00845020162 V or 8.4 mV

8 0

3 years ago

Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the

gregori [183]

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

Mostly at night would they have been seen.
Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

3 0

3 years ago

In a flying ski jump, the skier acquires a speed of 110 km/h by racing down a steep hill and then lifts off into the air from a

matrenka [14]

Answer:

Approximately $\displaystyle\rm \left[ \begin{array}{c}\rm191\; m\\\rm-191\; m\end{array}\right]$ .

Explanation:

Consider this $45^{\circ}$ slope and the trajectory of the skier in a cartesian plane. Since the problem is asking for the displacement vector relative to the point of "lift off", let that particular point be the origin $(0, 0)$ .

Assume that the skier is running in the positive x-direction. The line that represents the slope shall point downwards at $45^{\circ}$ to the x-axis. Since this slope is connected to the ramp, it should also go through the origin. Based on these conditions, this line should be represented as $y = -x$ .

Convert the initial speed of this diver to SI units:

$\displaystyle v = \rm 110\; km\cdot h^{-1} = 110 \times \frac{1}{3.6} = 30.556\; m\cdot s^{-1}$ .

The question assumes that the skier is in a free-fall motion. In other words, the skier travels with a constant horizontal velocity and accelerates downwards at $g$ ( $g \approx \rm -9.81\; m\cdot s^{-2}$ near the surface of the earth.) At $t$ seconds after the skier goes beyond the edge of the ramp, the position of the skier will be:

$x$ -coordinate: $30.556t$ meters (constant velocity;)
$y$ -coordinate: $\displaystyle -\frac{1}{2}g\cdot t^{2} = -\frac{9.81}{2}\cdot t^{2}$ meters (constant acceleration with an initial vertical velocity of zero.)

To eliminate $t$ from this expression, solve the equation between $t$ and $x$ for $t$ . That is: express $t$ as a function of $x$ .

$x = 30.556\;t\implies \displaystyle t = \frac{x}{30.556}$ .

Replace the $t$ in the equation of $y$ with this expression:

$\begin{aligned} y = &-\frac{9.81}{2}\cdot t^{2}\\ &= -\frac{9.81}{2} \cdot \left(\frac{x}{30.556}\right)^{2}\\&= -0.0052535\;x^{2}\end{aligned}$ .

Plot the two functions:

$y = -x$ ,
$\displaystyle y= -0.0052535\;x^{2}$ ,

and look for their intersection. Refer to the diagram attached.

Alternatively, equate the two expressions of $y$ (right-hand side of the equation, the part where $y$ is expressed as a function of $x$ .)

$-0.0052535\;x^{2} = -x$ ,

$\implies x = 190.35$ .

The value of $y$ can be found by evaluating either equation at this particular $x$ -value: $x = 190.35$ .

$y = -190.35$ .

The position vector of a point $(x, y)$ on a cartesian plane is $\displaystyle \left[\begin{array}{l}x \\ y\end{array}\right]$ . The coordinates of this skier is approximately $(190.35, -190.35)$ . The position vector of this skier will be $\displaystyle\rm \left[ \begin{array}{c}\rm191\\\rm-191\end{array}\right]$ . Keep in mind that both numbers in this vectors are in meters.