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lakkis [162]
3 years ago
7

Why are some consolation visible to New York State observers at midnight during April , but not visible at midnight during Octob

er
Physics
1 answer:
Svetach [21]3 years ago
3 0
"Midnight" means looking away from the Sun. But in 6 months from April to October the earth goes halfway around the Sun. So midnight in April and midnight in October are exactly opposite directions.
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Find the value of a x , the x-component of the object's acceleration.
jok3333 [9.3K]
To find the x component use the following formula, where Ф = theta = the angle 'a' makes with the x axis.

a_x = a*cos(theta)

3 0
3 years ago
8-30-2018 what phase change occurs as the substance at 0 degree c has its pressure dropped from 0.50 at to 0.25 atm
madam [21]

Answer:

at T = 0ºC the change of state is from the solid state to the gaseous state

Explanation:

In this exercise we are asked about the changes of state, from the data we will assume that the material is water.

Water can exist in three solid states, liquid and gas, in a graph of pressure ℗ against temperature (T) there is a point called triple at T = 0.01ºC, below this point the curve has two states at high pressure solid and low pressure gas.

As a result of the previous ones at T = 0ºC the change of state is from the solid state to the gaseous state

7 0
3 years ago
PLEASE HELP THIS IS DUE TODAY!! (GIVING BRAINLIEST TO WHOEVERY ANSWERS QUICKLY AND CORRECTLY!) 15 POINTS!
Jlenok [28]

As the core collapses, the outer layers of the star are expelled. A planetary nebula is formed by the outer layers. The core remains as a white dwarf and eventually cools to become a black dwarf. ... Like low-mass stars, high-mass stars are born in nebulae and evolve and live in the Main Sequence

hydrogen shell burning - outer layers swell. Red Giant Branch - helium ash core compresses - increased hydrogen shell burning. First Dredge Up - expanding atmosphere cools star - stirs carbon, nitrogen and oxygen upward - star heats up.

5 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
2 years ago
A bowling ball accidentally falls out of the cargo bay of an airliner as it flies along in a horizontal direction. As seen from
valina [46]

The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

<h3>Path of the bowling ball</h3>

Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

Learn more about horizontal direction here: brainly.com/question/2534565

#SPJ1

3 0
2 years ago
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