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pychu [463]
3 years ago
7

A particular star cluster contains stars all with the same apparent magnitude of +4. Near the cluster (and at the same distance

from the Earth) is a single star with an apparent magnitude of +1. The star's brightness appears to be the same as the collective brightness of the entire cluster.
How many stars are in this cluster?
Physics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:

3,000 stars.

Explanation:

Subtract the magnitude of the single star from the magnitude of the star cluster.

(+4) - (+1) = +3

Then multiply your answer by 1000

= 3,000 is a stars.

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Two positive charges of magnitude 20 micro C and 100 micro C are placed at a distance of 150cm. Calculate the force of repulsion
Luba_88 [7]

first \: positive \: charge = q1 = 20

1micro \: =   {1}^{ - 6}

1q = 20 \times 10 {}^{ - 6}

second \: charge = q2 = 100 \times 10 {}^{ - 6}

formula = f =  \frac{k \times q1 × q2}{d {}^{2} }

remember

k = 9 \times  {10}^{9}

change distance 150cm to 1.5m

putting values

f =\frac{9 \times 10 {}^{9} \times 20  \times 10 {}^{ - 6} \times 100 \times  {10}^{ - 6}   }{1.5 {}^{2} }

<h2 /><h3>answer </h3><h3><u>8</u><u>N</u></h3>
4 0
3 years ago
The design of a 60.0 cm industrial turntable requires that it has a kinetic energy of 0.250 j when turning at 45.0 rpm. What mus
Aneli [31]

Answer:

The moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

Explanation:

Given;

radius of the turnable, r = 60 cm = 0.6 m

rotational kinetic energy, E = 0.25 J

angular speed of the turnable, ω = 45 rpm

The rotational kinetic energy is given as;

E_{rot} = \frac{1}{2} I \omega ^2

where;

I is the moment of inertia about the axis of rotation

ω is the angular speed in rad/s

\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s

E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2

Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

5 0
3 years ago
A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N
Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

E_k = W = 40.2 j

mv^2/2 = 40.2

20v^2/2 = 40.2

10v^2 = 40.2

v^2 = 4.02

v = \sqrt{4.02} = 2 m/s  

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3 years ago
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MArishka [77]
The answer is “behaviorist”
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3 years ago
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Nastasia [14]
Sounds but not solids

hoped this helped  
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3 years ago
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