How many grams is 5kg of tap water?

1 answer:

3 0

When you write "5 kg", the little 'k' in the middle stands for 1,000.
So '5 kg' means '5,000 g'.

It doesn't matter what substance you're working with. It could be
tap-water, gold, air, or mud. 5 kg of it is still 5,000 grams of it.

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A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the

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Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

$D=\sqrt{9^2+12^2} \\\\D=15\ km$

For direction,

$\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}$

Hence, this is the required solution.

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What is the source of the radioactivity?

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The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1

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Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

$F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2})$ (1)

Where:

$F$ - Explosive force, measured in newtons.

$\Delta s$ - Barrel length, measured in meters.

$m$ - Mass of the shell, measured in kilograms.

$v_{o}$ , $v_{f}$ - Initial and final speeds of the shell, measured in meters per second.

If we know that $m = 1250\,kg$ , $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 750\,\frac{m}{s}$ and $\Delta s = 15\,m$ , then the explosive force experienced by the shell inside the barrel is: