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3 years ago

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Compared to a blue star in the same local cluster, the surface temperature of a red star is a) greater. b) the same. c) lower. d

) not consistently any of these

1 answer:

Andrew [12]3 years ago

3 0

Hello! My name is Zalgo and I am here to help you out on this concluding day. The answer would be C);lower. The reason it would be lower is because the hottest color of flames would be blue. Considering the way a start emits light is fire, this would be the most logical reason for it.

I hope that this helps! :P

"Stay Brainly and stay proud!" - Zalgo

(By the way, do you mind marking me as Brainliest? I'd greatly appreciate it! Thanks! X3)

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Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v

butalik [34]

Answer:

$u=34cm$

Explanation:

From the question we are told that:

Far point is $V=34 cm$

Near point is $u=17 cm$

Therefore

Focal Length

$f=-34cm$

Generally the equation for the Lens is mathematically given by

$\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$

$\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}$

$u=34cm$

5 0

2 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

3 years ago

Tom is throwing an baseball at an aluminum can,

pishuonlain [190]

Answer:

The question relates to the conservation of energy principle, the conservation of the linear momentum, and Newton's Laws of motion

Part A

1) Tom throwing a baseball at a can

The initial velocity of the baseball = v₂

The initial kinetic energy of the baseball, K.E.₂ = (1/2)·m₂·v₂²

∴ The final kinetic energy of the baseball, K.E.₂' = (1/2)·m₂·v₂'² < (1/2)·m₂·v₂²

Therefore, the energy of the ball before the collision is lesser than the energy of the ball after the collision

2) The evidence that would likely support the claim is that the baseball's height above the ground reduces rapidly immediately after the collision which is due to the reduced velocity, and therefore, the reduced (kinetic) energy

The final velocity of the baseball v₂' < v₂

Part B

1) The argument

The initial velocity of the can = v₁ = 0 (The can is initially at rest)

The initial kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² = 0

The final velocity of the can v₁' > v₁ = 0

∴ The final kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² > 0

Given that the velocity of the can increases from zero to a positive value after collision with the baseball, the kinetic energy of the can is increased from zero before the collision to a positive value after the collision

2) An evidence in support of the argument is the motion of the can which was initially at rest which is an indication of increase in energy podded by the can

Explanation:

8 0

3 years ago

A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2

lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

3 0

3 years ago

Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec

GaryK [48]

Mary and her younger brother Alex decide to ride the carousel at the State Fair, Mary's and Alex's angular speed M and tangential speed vM is mathematically given as

Mary's and Alex's angular speed=1.43

Tangential speed mary=3.22 m/s

Tangential speed alex =2.260m/s

<h3>What is Mary's and Alex's angular speed M and tangential speed vM?</h3>

Generally, the equation for angular speed is mathematically given as

$w=2\pi /T\\\\Therefore\\\\w=2\pi/3.9$

w = 1.61 rev/see 3.9

Centripetal acc mary = v^2/r

Centripetal acc mary = w^2r

Centripetal acc mary = w^2x 2m

Centripetal acc. of Alex = w²x L.u

Therefore

$\frac{(ac) mary }{(ac) plex}= 1.43$

Hence

tang. speed V=Wr

tang. speed of mary = 1.61x2 = 3.22 m/s

tang. speed of Alex: 1.61X1·4 =2.260m/s

Read more about Speed

brainly.com/question/4931057

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8 0

2 years ago