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2 years ago

Choose all the answers that apply.

Physics

2 answers:

DerKrebs [107]2 years ago

7 0

Answer:

use their own matter to create nuclear reactions

generate their own energy

are classified primarily by temperature

usually seem bright if they are close to Earth

Explanation:

kvv77 [185]2 years ago

4 0

All of the answers are correct !!

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What is 6,210 bucks converted to kilobucks?

Vinvika [58]

When 6210 bucks is converted to kilobucks, the result obtained is 6.21 kilobucks

<h3>Conversion scale </h3>

1000 bucks = 1 Kilobuck

Using the above convesion scale, we can express 6210 bucks in kilobucks

<h3>How to express bucks in kilobucks</h3>

1000 bucks = 1 Kilobuck

Therefore,

6210 bucks = (6210 × 1) / 1000

6210 bucks = 6.21 kilobucks

Thus, 6210 bucks is equivalent to 6.21 kilobucks

Learn more about conversion:

brainly.com/question/2139943

4 0

2 years ago

Read 2 more answers

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

What is the wavelength associated with an electron with a velocity of 4.8X10s m/s? (Mass of the electron is 9.1X10-31kg)?

amid [387]

Answer:

1.52 nm

Explanation:

Using the De Broglie wavelength equation,

λ = h/p where λ = wavelength associated with electron, h = Planck's constant = 6.63 × 10⁻³⁴ Js and p = momentum of electron = mv where m = mass of electron = 9.1 × 10⁻³¹ kg and v = velocity of electron = 4.8 × 10⁵ m/s

So, λ = h/p

λ = h/mv

substituting the values of the variables into the equation, we have

λ = h/mv

λ = 6.63 × 10⁻³⁴ Js/(9.1 × 10⁻³¹ kg × 4.8 × 10⁵ m/s)

λ = 6.63 × 10⁻³⁴ Js/(43.68 × 10⁻²⁶ kgm/s)

λ = 0.1518 × 10⁻⁸ m

λ = 1.518 × 10⁻⁹ m

λ = 1.518 nm

λ ≅ 1.52 nm

4 0

3 years ago

Which property increases as an electromagnetic wave's energy decreases?

aleksklad [387]

The answer is B frequency. When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases

4 0

3 years ago

Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.