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2 years ago

What is the apparent magnitude of the brightest star in the Big Dipper?

2 answers:

4 0

The magnitude of Alioth ( the brightest star in the big dipper ) is 1.76 and it is about 81 light years distant from Earth.

ollegr [7]2 years ago

3 0

Answer: +1.77

Explanation:

Big dipper is the asterism made of seven bright stars in Ursa Major constellation. An asterism a pattern formed by joining the bright stars in the night sky. On the other hand, whole night sky is divided into 88 areas called constellations. These asterisms are used to identify the constellations.

The brightest star of the big dipper is Alioth which is the 32nd brightest star in the night sky. Its apparent magnitude is +1.77.

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melamori03 [73]

All three windows are the same size.

A has 10 complete waves visible through the window. B has 3, and C has 4.

So A must have the smallest wavelengths.

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3 years ago

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An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remai

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3 years ago

A 1.2 kg block of wood hangs motionless from strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes emb

Firdavs [7]

Answer:

speed of the bullet before it hit the block is 200 m/s

Explanation:

given data

mass of block m1 = 1.2 kg

mass of bullet m2 = 50 gram = 0.05 kg

combine speed V= 8.0 m/s

to find out

speed of the bullet before it hit the block

solution

we will apply here conservation of momentum that is

m1 × v1 + m2 × v2 = M × V .............1

here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet

put all value in equation 1

m1 × v1 + m2 × v2 = M × V

1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8

solve it we get

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3 years ago

You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp

GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(d) $539.55\mu$

(e) 0

Explanation:

The period for 1 circle $2\pi$ of the merry go around is 9.5s. It means the angular speed is:

$\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s$

(a)The speed is

$v = \omega * R = 0.661 * 17 = 11.24 m/s$

(b) Centripetal acceleration:

$a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2$

$F = ma = 55 * 7.44 = 409 N$

(d) let the coefficient of friction by $\mu$ . The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

$F_f = mg\mu = 55*9.81\mu = 539.55\mu$

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2 years ago

According to Newton's Second Law, the force of the club hitting the golf ball will cause it to accelerate. At the moment of impa

vazorg [7]

Answer:

Option B

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3 years ago

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