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3 years ago

(NEED HELP! +28pts and BRAINLIEST) The total energy output of a star per second is called ________. Select all that apply.

Physics

2 answers:

topjm [15]3 years ago

8 0

Answer:

The answer would be B.) Luminosity.

Explanation:

The luminosity of a star is basically, as the question says, the total energy output of a star per second. The luminosity is also the brightness of a star. If a star has a lot of energy, the star would be bright.

Alekssandra [29.7K]3 years ago

3 0

Answer:

luminosity

Explanation:

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The Amazon rainforest has millions of trees down in South America. Pantanal, the largest wetland, or swamp, on Earth sits direct

Vikentia [17]

Answer:The Pantanal is, when the season changes of the river flow, the pantanal shrinks after a few river channels The amazon rainforest has a lot of trees.The vapor rises and condenses it into clouds.Water leaves the forest and comes back, the amazon leaves itself.If the amazon water was not to go by itself everything would be different.

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2 years ago

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Suppose that, from measurements in a microscope, you determine that a certain bacterium covers an area of 1.50 μm2. Convert this

Paladinen [302]

1 m = 1 000 000 ym

converted other way we can say that:

1 ym = $10^{-6}$ m

Now, since we have ym^2 which is ym*ym which means:
1 ym^2 = $(10^{-6}) ^2 = 10^{-12}$ m

we have 1,5 ym^2 which means that answer is:
$1.5* 10^{-12} m$

8 0

3 years ago

WILL MARK BRAINLIEST AND THANK

timurjin [86]

Same temperature and difference in air pressure

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3 years ago

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uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

Please help! Will give brainliest. 10 points. Show work!

Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

5 0

3 years ago