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3 years ago

An star system is made of two stars.

Physics

1 answer:

guapka [62]3 years ago

7 0

Answer:

A star system is made of two stars.

Explanation:

A star system is a system made up of a few number of stars orbiting each other, and bound by gravitational attraction. It is also known as a 'stellar system', and can be made up of two, or more stars. The star system composed of two stars is known as a Binary star system, and both stars circle the center of mass of their system. Multiple star system composed of three or more stars also exists, although rarely, and the stars in such systems are bound by gravitational attraction, although distance between them may or may not be close. A very large group of more than a hundred to a thousand cluster of stars is known as a 'Open Star Cluster.'

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An arthroscope can be inserted into a knee joint through a very tiny incision to help a doctor examine the knee. A light is tran

natali 33 [55]

Sample Response: The technology being described is fiber optics. Since they are small and flexible, they enable doctors to see areas that they might otherwise be unable to see without surgery.

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3 years ago

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the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a cur

cricket20 [7]

Answer:

0.03605 V/m is the electric field in the gold wire.

Explanation:

Resistivity of the gold = $\rho = 2.44\times 10^{-8} \Omega.m$

Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

Diameter of the wire = d = 0.9 mm

Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = $0.45\times 0.001 m$

( 1mm = 0.001 m)

Area of the cross-section = $A=\pi r^2=\pi r^(0.45\times 0.001 m)^2$

Resistance of the wire = R

Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

$R=\rho\times \frac{L}{A}$

$V(voltage)=I(current)\times R(Resistance)$ ( Ohm's law)

$\frac{V}{I}=\rho\times \frac{L}{A}$

We know, Electric field is given by :

$E=\frac{dV}{dr}$

$E=\frac{V}{L}$

$E=\frac{V}{L}=\rho\times \frac{I}{A}$

$E=2.44\times 10^{-8} \Omega.m\times \frac{0.940 A}{\pi r^(0.45\times 0.001 m)^2}=0.03605 V/m$

0.03605 V/m is the electric field in the gold wire.

3 0

4 years ago

A force is required for any change in an object's motion to occur. True False

sertanlavr [38]

Answer:

True

Explanation:

an object in motion stays in motion unless acted upon by another force

7 0

3 years ago

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Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

IgorC [24]

Answer:

The answers to the questions are;

a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.

b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.

c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.

Explanation:

To solve the question, we write out the known variables as follows

1 g of fat = 9.00kcal

Number of steps the student climbs = 95 steps

Height of each step = 0.150 m

Time it takes for the student to reach the top of the stairs = 57.5 s.

Efficiency of human muscles = 20 %

Mass of student, m = 65 kg

a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.

The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..

That is work done, W, = P. E. = m·g·h

Where:

h = The total height climbed by the student

g = Acceleration due to gravity = 9.81 m/s²

Therefore;

h = Height of each step × Number of steps the student climbs =

= 0.150 m/(step) × 95 steps = 14.25 m

Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²

= 9086.5125 J

We remember that the efficiency of the muscle is 20 %

The formula for efficiency is

Efficiency = $\frac{Ene rgy Out put}{Energ y In put} \times 100 %$

The work produced by the muscle = Energy Output = 9086.5125 J

Energy input is given by

$\frac{Out put} {Effici ency}$ = 9086.5125 J/ (0.2) = 45432.5625 J

= 45.432 kJ

From the question, 1 g of fat = 9.00 kcal and

1 kcal = 4186 J

Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J

Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ

To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs $\frac{37674 kJ}{45.432 kJ}$ times to make up the 37674 kJ energy contained in 1 kg of fat