All categories

3 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]3 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.

You might be interested in

slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl

11111nata11111 [884]

Answer:

$v_{s}=65.2km/h$

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

$t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0$

Apply quadratic formula to solve for vs

$v_{s}=65.2km/h$

4 0

3 years ago

At the presentation ceremony, a championship bowler is presented a 1.60-kg trophy which he holds at arm's length, a distance of

nikdorinn [45]

Answer:

a)10.28 Nm

b)9.93 Nm

Explanation:

Let g = 9.81m/s2. First we can calculate the weight of the trophy

W = mg = 1.6 * 9.81 = 15.696 N

(a) The torque is product of force and its moment arm

T = WL = 15.696 * 0.655 = 10.28 Nm

(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:

$L_2 = Lcos(15^0) = 0.655cos(15^0) = 0.655*0.966 = 0.633 m$

Again, torque is product of force and its moment arm

$T_2 = WL_2 = 15.696 * 0.633 = 9.93 Nm$

8 0

3 years ago

A cows mass is 401 kg and a trucks mass is 832 kg. What is the difference between their weights . Answer please!!!

jeyben [28]

Answer:

403 kg is the diffrence

4 0

3 years ago

How long does it take for light to get to earth from the moon?

FrozenT [24]

About 1.3 SECONDS.............

4 0

3 years ago

If the heat of fusion is 32.2 kj/mol, the amount of heat energy required to melt 5.67 grams of feo is

anzhelika [568]

The heat of fusion is the amount of energy required to change the state of one mole of substance from solid to liquid at constant temperature. Therefore, in order to determine the energy needed by a given mass of substance, we must first convert the mass into moles by using:
Moles = mass / Mr
Moles = 5.67 / 71.84
Moles = 0.079 mol

Energy = latent heat of fusion * moles
Energy = 32.2 * 0.079
Energy = 2.544 kJ = 2,544 J

3 0

3 years ago