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2 years ago

Can someone help answer question 8?

Physics

2 answers:

tigry1 [53]2 years ago

5 0

Star A - (color) red-orange (type) main sequence

Star B - (color) yellow (type) supergiant

Star C - (color) blue-white (type) white dwarf

Star D - (color) yellow (type) main sequence

*im pretty sure these are correct if I remember clearly*

Liono4ka [1.6K]2 years ago

4 0

Answer:

A- Orange-main sequences

B- Red-red supergiant

C- White-White dwarf

D- Yellow-Red giant

Explanation: there's picture on the internet

'star types chart' 'star types and sizes'

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How far away from the sun would a planet mercury's size have to be before it would have an atmosphere?

Elza [17]

It's 87.97 days of the orbital period going round the sun it's days of the shortest of all planets in the solar system.

4 0

3 years ago

Give short definitions for the following terms: 1. All Atomic number 2. Mass number 3. Atomic weight 4. Isotope 5. Natural abund

son4ous [18]

Explanation:

Hello ! Let's solve this!

1-atomic number: the atomic number is the number of protons that the atom has

2-mass number: is the amount of protons plus the amount of neutrons in the atom

3-atomic weight: average of the masses of an atom per 1/12 of the mass of carbon

4-isotope: it is the same element that has the same number of protons but a different number of neutrons

5-natural abundance: measures the amount of isotopes of an element in nature

6-unit of atomic mass: (uma) is a unit that is equivalent to 1/12 of the carbon atom

8 0

3 years ago

A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t

Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

$\omega= 19 rev/s$

we need to convert it into radiands per second. We know that

$1 rev = 2 \pi rad$

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

$\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s$

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

$\theta= \omega t$

where

$\omega=119.3 rad$

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

$\theta=(119.3 rad/s)(5 s)=596.5 rad$

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

$\alpha = 1.6 rad/s^2$

So the final angular speed will be given by

$\omega_f = \omega_i + \alpha \Delta t$

where

$\omega_i = 119.3 rad/s$ is the initial angular speed

$\alpha = 1.6 rad/s^2$ is the angular acceleration

$\Delta t = 15 s - 10 s = 5 s$ is the time interval

Solving the equation,

$\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s$

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

$\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2$

Substituting the numbers into the equation, we find

$\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad$

(e) 222 m

The instantaneous speed of the center of the wheel is given by

$v_{CM} = \omega R$ (1)

where

$\omega$ is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

$\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s$

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

$v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s$

And so the displacement of the center of the wheel will be

$d=v_{CM} t = (44.4 m/s)(5 s)=222 m$

8 0

3 years ago

Which of the following is transferred in order for static electricity to occur?

Svetllana [295]

Electrons is transferred in order for static electricity to occur.

6 0

3 years ago

Read 2 more answers

Garrick rubs an inflated balloon against his hair. He then touches the balloon against a non-conducting wall.

Sauron [17]

Answer:

Figure A

Explanation:

At first, the inflated balloon is rubbed against the hair.

In this situation, the balloon is charged by friction: because of the friction between the surface of the balllon and the hair, electrons are transferred from the hair to the surface of the balloon.

As a result, when the balloon is detached from the hair, it will have an excess of negative charge (due to the acquired electrons).

Then, the balloon is placed in contact with the non-conducting wall.

The non-conducting wall is initially neutral (equal number of positive and negative charges).

Because the wall is made of a non-conducting material (=isolant), the charges cannot move easily through it. Therefore, even though the charges on the wall feel a force due to the presence of the electrons in the balloon, they will not redistribute along the wall.

Therefore, the charges on the wall will remain equally distributed, as shown in figure A.

7 0

3 years ago