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Sphinxa [80]
2 years ago
5

Why doesn't the moon turn black during a total lunar eclipse? Explain your answer​

Physics
1 answer:
Alisiya [41]2 years ago
4 0

Answer:

Thank me mark me brainliest

Explanation:

I did not explain because i got a photo

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A restaurant records the number of tables served each night, and the results have the values: minimum = 3, lower quartile = 14,
dybincka [34]

To choose the correct box plot, verify each of the options and make sure all the values in the plot match the values provided.

<h3>How to identify the median?</h3>

In a box plot, this value is represented by a vertical line located in the middle of the graph.

<h3>How to identify the maximum and the minimum?</h3>

The maximum is the value located on the farthest right, while the minimum is located on the farthest left.

<h3>How to identify the quartiles?</h3>

Divide the graph into 4 and analyze how much each quartile represents.

Learn more about graphs in: brainly.com/question/16608196

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5 0
2 years ago
Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False
kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

\vec{L} = I \vec{\omega}

where I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2 is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2} so \vec{\omega} \neq 0 unless \vec{r} ║  \vec{v}.

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0
2 years ago
A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0
3 years ago
How is velocity and instantaneous speed alike
allsm [11]
<span>velocity is defined as the rate of change of displacement irrespective of the length of the path travelled while speed is the average rate of covering distance. but in the liming case where the instantaneous velocity is given as v=dx/dt where dx is the small displacement in a small interval dt, both the speed and velocity have the same magnitude and the direction of velocity is the direction of the tangent to the corresponding displacement-time curve.</span>
5 0
3 years ago
Determine the torque<br> produced by a perpendicular force of 75<br> N at the end of a 0.2 m wrench.
trasher [3.6K]

Answer:c

Explanation:

Because I took uty

7 0
2 years ago
Read 2 more answers
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