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4 years ago

Average atomic must be

Physics

2 answers:

Zigmanuir [339]4 years ago

7 0

Answer:

MASS

Explanation:

I took the quiz

omeli [17]4 years ago

6 0

I’m not quit sure sorry?

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During the storm a tree fell over into a river what might happen to the tree

mart [117]

The tree might get swept away by the current and it will disappear when it catches on something

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3 years ago

1.14 Which of the following is an example of a force without touching? A A boy pushing a trolley. B The mass of a car. C с A mag

andriy [413]

Explanation:

Option C a magnet attracting a nail

4 0

3 years ago

Why do different substances have different melting and boiling points?

azamat

Because the atoms and molecules all have different properties

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3 years ago

A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C

mario62 [17]

Answer:

3.62 m and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}$

$\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}$

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

$\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}$

$\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}$

x = - 1.4 m

6 0

4 years ago

Will mark the brainliest

AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

$J = \Delta p = p_{f} - p_{i}$

Where:

$p_{f}$ : is the final momentum

$p_{i}$ : is the initial momentum

The initial momentum is given by:

$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

$v_{1_{i}} = -10 m/s$

And because the nail is not moving:

$v_{2_{i}}= 0$

$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

$p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s$

Finally, the impulse is:

$J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s$

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!

7 0

3 years ago