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Anuta_ua [19.1K]
3 years ago
12

A1.200 kg car is sliding down an icy

Physics
1 answer:
Triss [41]3 years ago
5 0

Answer:

1585.67N

Explanation:

their are many student who can not get answer on time and step by step answer.so there are

a wats up group of experienced experts who helps you for free with step by step answer.just join this and get answer.

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Which term best describes the basic unit that makes up all matter?
neonofarm [45]

Answer:

C. Atoms

Explanation:

The basic unit of matter and the smallest, indivisible unit of a chemical element. It comprises a nucleus (neutrons + protons) that is surrounded by a cloud of electrons.

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2 years ago
Side effects such as drowsiness, mental clouding, and nausea can
Mila [183]
Tranquilizers

Because they are meant to put you at ease which may cause drowsiness and mental clouding
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What is the power of a diverging lens of focal length 0.4 m
romanna [79]

Answer:

-2.5 is the answer to your question

4 0
2 years ago
Inertia law what happens when You are standing on the bus and the bus stops abruptly:
I am Lyosha [343]

Answer:

You will fly forward in the bus until you hit something.

Explanation:

While standing there on the bus, you are traveling at the same speed as the bus. If the bus suddenly stops, you will still be traveling at the same speed you started with. That is until you hit something hard enough or big enough to stop you.

4 0
3 years ago
Read 2 more answers
Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o
Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, Q_1 and Q_2, separated by a distance, d, is given

F = k\dfrac{Q_1Q_2}{r^2}

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

F_X = k\dfrac{18Q}{x^2}

Force due to charge Y

F_Y = k\dfrac{-27Q}{(0.60-x)^2}

Since both forces are equal and opposite,

F_X = -F_Y

k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}

\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}

2(0.60-x)^2 = 3x^2

2(0.36-1.20x+x^2) = 3x^2

0.72-2.40x+2x^2 = 3x^2

x^2+2.40x-0.72 = 0

Applying the quadratic formula,

x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}

x = -2.7 or x = 0.27

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0
3 years ago
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