Which model shows how an image appears to a person when they look through a concave lens?

AS AIRPLANE WENT FROM 120 m/s to 180 m/s in 4.0 seconds what is the acceleration

marusya05 [52]

Answer:15metre per second squ

I

Explanation:

acceleration=(final velocity-initial velocity)÷t

acceleration=(180-120)÷4

acceleration=60÷4

acceleration=15 metre per second square

8 0

3 years ago

Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second

zlopas [31]

Given:

10^10 electrons per second

To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:

From literature,

1 Coulomb is equivalent to 6.242×10^18 electrons.

So,

= 10^10 electrons * (1 coulomb/6.242×10^18 electrons) / second
= 1.602 x 10^-9 coulumbs

This value is too small to be used in an actual setting.

3 0

3 years ago

Arrange the core steps of the scientific method in sequential order.

andriy [413]

the picture bellow gives you all the answers, this picture was given to my class by my science teacher a few years ago so it is accurate

4 0

3 years ago

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A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo

Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

$W = \Delta U = U_2 - U_1$

where the potential energy is defined as follows

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}$

Let's first calculate the distance 'r' for both positions.

$r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m$

Now, we can calculate the potential energies for both positions.

$U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J$

Finally, the total work done on the moving particle can be calculated.

$W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J$

4 0

3 years ago

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Which best explains satellites? the moon is a satellite of the earth, and the earth is a satellite of the sun. mercury and pluto

Basile [38]

the moon is a satellite of the earth, and the earth is a satellite of the sun.
that is the best answer

6 0

3 years ago

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