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2 years ago

The star named Canopus has a declination of approximately –52°. Which of these statements is correct about Canopus?

1 answer:

6 0

It’s A, it’s fifty two degrees above the celestial equator, you can also look up the same question and see if any of them match

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(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel

sammy [17]

Answer:

Part a)

$R_1 = 0.072 m$

Part b)

$R_2 = 0.036 m$

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

$R = \frac{mv}{qB}$

now for single ionized we have

$R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}$

$R_1 = 0.072 m$

Part b)

Similarly for doubly ionized ion we will have the same equation

$R = \frac{mv}{qB}$

$R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}$

$R_2 = 0.036 m$

Part c)

The distance between the two particles are half of the loop will be given as

$d = 2(R_1 - R_2)$

$d = 2(0.072 - 0.036)$

$d = 0.072 m$

6 0

2 years ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice the

saul85 [17]

Answer:

The wire now has less (the half resistance) than before.

Explanation:

The resistance in a wire is calculated as:

$R=\alpha \frac{l}{s}$

Were:

R is resistance

$\alpha$ is the resistance coefficient

l is the length of the material

s is the area of the transversal wire, in the case of wire will be circular area ( $s=\pi r^{2}$ ).

So if the lenght and radius are doubled, the equation goes as follows:

$R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }$

So finally because the circular area is a square function, the resulting equation is half of the one before.

7 0

3 years ago

Who agrees to pay for certain types of losses in exchange for payments on a policy?

Vilka [71]

Passes the the law then through the people where it is followed

8 0

3 years ago

Read 2 more answers

Terry dropped a pebble from the top of his apartment building. The pebble impacted the ground after falling for 2.2 seconds. Cal

Contact [7]

The final velocity of the pebble as it hits the ground is 21.5 m/s.

We have to use the equation;

v = u + gt

Where;

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Since the pebble was dropped from a height, u =0 m/s, hence,

v = gt

v = 9.8 ms-2 × 2.2 s

v = 21.5 m/s

Learn more about velocity: brainly.com/question/8898885

6 0

2 years ago

Calculate the power in kilowatts required to lift a 500 kg mass 12 meters in one second

ivanzaharov [21]

The weight of the load is (mass) x (gravity).

The work done to lift it is (weight) x (distance) = (mass) x (gravity) x (distance)

The power is (work)/(time) =

   (mass) x (gravity) x (distance) / (time) =

   (500 kg) x (9.8 m/s²) x (12 m) / (1 sec) =

(500 x 9.8 x 12 / 1) (kg-m / sec²) (m) / (sec) =

   (500 x 9.8 x 12 / 1) newton - meter / sec =

(500 x 9.8 x 12 / 1) joule    / sec =

   58,800 watts

   = <em>58.8 kW</em>

   =    about 78.8 horsepower

8 0

2 years ago