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bixtya [17]
3 years ago
9

Planet-solar system-galaxy-local group. what comes next

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0

Answer:

Super cluster

Explanation:

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Electromagnetism consists of what two processes?
kherson [118]

Answer: it’s A and B

Explanation: everyone else on this post was giving you the wrong answer.

5 0
3 years ago
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What would it mean if neither the red or blue litmus paper changes colors?
Anna35 [415]

if the color changes, it is neutral but if it stays the same, it is an acid.

8 0
3 years ago
How much heat is necessary to warm 500g of water from 20°c to 65°c​
Vsevolod [243]

Explanation:

so sorry

don't know but please mark me as brainliest please

8 0
3 years ago
A semicircular plate with radius 6 m is submerged vertically in water so that the top is 2 m above the surface. Express the hydr
dalvyx [7]

Answer: 313920

Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.

Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.

The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.

The depth below the water surface of each strip is,

di = 8 − yi

and that in turn gives us the pressure on the strip,

Pi =ρgdi = 9810 (8−yi)

The area of each strip is,

Ai = 2√4− (yi) 2Δy

The hydrostatic force on each strip is,

Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy

The total force on the plate is found on the attached image.

5 0
3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
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