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yarga [219]
3 years ago
11

Which of these best describes how an appropriate star chart is selected to locate objects in the sky?

Physics
2 answers:
lisov135 [29]3 years ago
7 0

Answer:

match the day and time of the year and location

Explanation:

The stars that we see on the sky change with the station of the year, they change their position depending of the day of the year and the location, mainly the latitude since on that depends what "side" of the earth the sky is facing towards, at certain points the north hemisphere sees a different sky than the south, but the east and west as long as they are in the same latitude do see the same sky.

Sergio [31]3 years ago
6 0
The layout of the stars in the sky is determined by the date, time of night, and your location (mainly latitude). So to pick the best star chart, you should go with the one that's closest to the present date and your location, then make allowance for what time it is. Everything in the sky moves about a degree every 4 minutes.
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Why do we feel the effects of gravity given that the gravitational force is so weak?
almond37 [142]
Any two objects in the universe attract each other. Gravity is the force exerted by earth on you (you exert the same force on earth) but due to the fact that earth has a huge mass compared to yours, you will be attracted to earth only by a small gravitational force.

6 0
3 years ago
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A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi
Olenka [21]
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J
8 0
2 years ago
Which soil horizon is located closest to the earth's crust?
Pachacha [2.7K]

Answer: O horizon

Horizons refers to the distinct layers of soil lying parallel to the earth surface. Horizons develop as a result of soil formation. Soil forms as a result of weathering or rocks and addition of organic matter from the decomposition of plant and animal waste. Each horizon differs from the others on the basis of color, texture, type of particles present in the soil, type of minerals present and amount of organic matter present in the soil.

O horizon is the soil horizon that is located closest to the earth's crust. This horizon consist of undecayed or partially decayed animal and plant waste like shedded leaves, bark, animal skin and feces. As, the matter remains undecomposed, therefore, this horizon consists of low amount of organic matter and it is less fertile for plant growth.  

7 0
3 years ago
Read 2 more answers
(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14
Elenna [48]

Answer:

w = 5832.372 Joules

Explanation:

Mass of water, m = 20 kg

The water was pulled up to a height of 35 meters, i.e. h = 35 m

It takes 14 minutes to pull up the water through the height, 35 m

speed = distance/ time = 35/14 = 2.5 m/min

The bucket's height, y = speed * time = 2.5t meters

6 kg of water drips out of the bucket throughout the 14 minutes

The rate at which the water drips drips out = (6/14) = 0.4286 kg/min

Mass of water that drips out in time, t = 0.4286t kg

The mass of water remaining = (20 - 0.4286t) kg

Change in Workdone, Δw = mgΔy

Δy = 2.5 Δt

Δw = mg *  2.5 Δt

dw =  (20 - 0.4286t)g2.5 dt

integrating both sides

dw = (50g - 1.07gt)dt

w = \int\limits^a_b {(50g-1.07gt)} \, dx where b = 0, a = 14

w = 50gt - 1.07g(t²)/2      g = 9.8 m/s²

w = 490t - 5.243t²

w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)

w = 6860 - 1027.628

w = 5832.372 Joules

3 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
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