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GrogVix [38]
3 years ago
5

Draw the distance time graph for a body at rest​

Physics
2 answers:
Semenov [28]3 years ago
8 0

I hope it is helpful for you...

cupoosta [38]3 years ago
3 0
Just a straight line cuz the body is not moving
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On earth, what is a child’s mass if the force of gravity on the child’s body is 100 N
ikadub [295]

Answer: 10.2 kg if g = 9.8, 10 if g = 10.

Explanation:

Weight or the "force of gravity" on a person is simply defined by the equation: F = ma. In this case, the acceleration is g, which is 9.8 but can be rounded up to 10. Based on this, we have:

F = mg

100 = m*9.8

m = 10.2(or 10 if we set g to 10).

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3 years ago
if you push a small act with 7,000 N force and push a heavy truck with the same force, which one will have a greater acceleratio
inn [45]

Answer:

the small one will have greater acceleration

Explanation:

force and mass are inversely proportional. force and acceleration are directly proportional. which means greater mass have smaller acceleration and smaller mass has greater acceleration. this is due to newtons second law of motion.

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3 years ago
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The decomposition of water into hydrogen gas H2 and oxygen gas O2 can be modeled by the balanced chemical equation
Volgvan

2H2O->2H2+O2


This balanced chemical equation represents the decomposition of water into hydrogen gas and oxygen gas



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3 years ago
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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

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4 years ago
Choose the correct statement about the acceleration of the car.
brilliants [131]

Answer:

Whats the question/word problem or where is the graph (if included) representing this problem?

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3 years ago
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