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3 years ago

8

How many cubic centimeters of area does 762 g of distilled water occupy?

1 answer:

BartSMP [9]3 years ago

8 0

1 kg of water = 1 L = 1 dm³
762 g of water = 762 cm³

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Four solid balls, each with a different mass, are moving at the same speed. Which ball would require the most force to stop its

zhannawk [14.2K]

Obviouly 20kg.............

8 0

4 years ago

Read 2 more answers

How can water boil without heat?

kaheart [24]

Answer:

Put water at room temperature into a vacuum chamber and begin removing the air. Eventually, the boiling temperature will fall below the water temperature and boiling will begin without heating. Or if you want to be easy but messy, add dry ice to a bowl of water and watch how the water starts to boil.

3 0

3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force $N=W-F$ . On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

7 0

3 years ago

Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile? Answ

Vinil7 [7]

Answer: 272.82 drop/tile

Explanation:

Given that the Rain drops fall on a tile surface at a density of 4638 drops/ft2. There are 17 tiles/ft2. How many drops fall on each tile?

Tiles/ft^2 × drop/tiles = drop/ft^2

Tiles will cancel out. Leaving the answer to be drop/ ft^2

Substitutes all the magnitude of the above units.

17 × drop/tiles = 4638

Make drop/tiles the subject of formula

Drop/tiles = 4638/17

Drop/tiles = 272.82

Therefore, 272.82 drop/tile drops fall on each tile?

8 0

3 years ago

A nichrome wire and an aluminum wire, each with the same initial resistance, have the same change in resistance when heated sepa

Svetach [21]

Answer:

The ratio of temperature change of nichrome wire to the temperature change of aluminum wire is found to be <u>9.75</u>

Explanation:

The change in resistances and initial resistances of the wires are:

Change in Resistance of Nichrome Wire = ΔR₁

Change in Resistance of Aluminum Wire = ΔR₂

Initial Resistance of Nichrome Wire = R₁

Initial Resistance of Aluminum Wire = R₂

Also, it is given that the change in resistance and initial resistance of both wires is equal. Therefore,

ΔR₁ = ΔR₂ ---------- eqn (1)

R₁ = R₂ ---------- eqn (2)

The change in Resistance due to temperature is given by formula:

ΔR = R α ΔT

Therefore, eqn (1) becomes:

R₁ α₁ ΔT₁ = R₂ α₂ ΔT₂

using eqn (2):

α₁ ΔT₁ = α₂ ΔT₂

ΔT₁/ΔT₂ = α₂/α₁

where,

α₁ = Temperature coefficient of resistance of nichrome = 0.4 x 10⁻³ °C⁻¹

α₂ = Temperature coefficient of resistance of aluminum = 3.9 x 10⁻³ °C⁻¹

Therefore,

ΔT₁/ΔT₂ = (3.9 x 10⁻³ °C⁻¹)/(0.4 x 10⁻³ °C⁻¹)

<u>ΔT₁/ΔT₂ = 9.75</u>

6 0

4 years ago