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3 years ago

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70 POINTS AND BRAINLIEST PLEASE HELP!!!The Moon orbits Earth. This orbit causes the Moon to look different over the course of ab

out a month. These differences are called phases. One phase is called a full moon. Look closely at the table to the right. What can you conclude about a full moon? A. A. A full moon only occurs in the summer months. B. B. A full moon always occurs after a first quarter moon. C. C. A full moon always occurs on the 5th day of the month. D. D. A full moon always occurs before the first quarter moon.

1 answer:

Rudik [331]3 years ago

8 0

Answer:

It always acurs after a 1st quarter

do you have photo?

Explanation:

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A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from t

dalvyx [7]

Answer:

The net torque about the pivot is and the answer is 'c'

c. $T_{net}=8.58$

Explanation:

$T=F*d$

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

$T_1=F_1*d_1$

$T_1=7.8N*1.6m=12.48N*m$

$T_2=F_2*d_2$

$T_2=2.60N**cos(30)*3.0m$

$T_2= - 3.9 N*m$

Now to find the net Torque is the summation of both torques

$T_{net}=T_1+T_2$

$T_{net}=12.48N-3.9N=8.58N$

8 0

3 years ago

How many protons are in this atom if it has a balanced charge

Lerok [7]

It would be 20 protons left because 1-19= 20

7 0

2 years ago

A 1.00 kg particle has the xy coordinates (-1.20 m, 0.500 m) and a 4.50 kg particle has the xy coordinates (0.600 m, -0.750 m).

just olya [345]

Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

Explanation:

The center of mass of a system of particles ( $\vec r_{cm}$ ), measured in meters, is defined by this weighted average:

$\vec r_{cm} = \frac{\Sigma_{i=1}^{n}\,m_{i}\cdot \vec r_{i}}{\Sigma_{i=1}^{n}\,m_{i}}$ (1)

Where:

$m_{i}$ - Mass of the i-th particle, measured in kilograms.

$\vec r_{i}$ - Location of the i-th particle with respect to origin, measured in meters.

If we know that $\vec r_{cm} = (-0.500\,m,-0.700\,m)$ , $m_{1} = 1\,kg$ , $\vec r_{1} = (-1.20\,m, 0.500\,m)$ , $m_{2} = 4.50\,kg$ , $\vec r_{2} = (0.600\,m, -0.750\,m)$ and $m_{3} = 4\,kg$ , then the coordinates of the third particle are:

$(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}$

$(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})$

$(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)$

$(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)$

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

5 0

2 years ago

Zoe finds that the temperature of a substance is 12 degrees Celsius. What does this tell Zoe about the substance?

maks197457 [2]

Given the temperature, we can tell if the substance is cold or not relative to the reference temperature. For example, compared to the substance having a temperature of 15 degrees C, the substance is colder and it is hotter from the substance of temperature lesser than 12 degrees C.

5 0

2 years ago

Read 2 more answers

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest

shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer: 3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f = 5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

5 0

3 years ago