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1 year ago

Part A

Physics

1 answer:

Basile [38]1 year ago

6 0

The relationships to determine the number of calories to change 0.50 kg of 0°C ice to 0°C ice water is 1,080,000 cal.

<h3>How does heating ice that is at C affect it?</h3>

Ice melts and becomes liquid water at 0 degrees Celsius. Once all of the ice has been entirely transformed into liquid water, the temperature of the remaining ice begins to increase once more (in °C), continuing to rise until it reaches 100 °C, where it then stabilizes.

The water turns into steam when it reaches a temperature of 100 °C (D).

Water has a fusion latent heat of fusion of 80 cal/g.

Water has a 1 cal/g-C specific heat.

Water has a 540 cal/g latent heat of vaporization.

In light of this, the total amount of heat needed is 1500 g [(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g [(720 cal/g)] = 1,080,000 cal.

To learn more about Vaporization refer to:

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Two charges, - Q0 and - 4Q0 are a distance d apart. These two charges are free to move but do not because there is a third charg

TEA [102]

To solve this problem we will use the equilibrium conditions in the Electrostatic Forces. In turn, we will use the concept formulated from Coulomb's laws to determine the intensity of the Forces and make the respective considerations.

Our values for the two charges are:

$q_1 = -Q_0$

$q_2 = -4Q_0$

As a general consideration we will start by determining that they are at a unit distance (1) separated from each other. And considering that both are negative charges, they will be subjected to repulsive force. Said equilibrium compensation will be achieved only by placing a third force between the two.

Let the third charge be $q_3 = +Q$ is placed at a distance x from $q_1$

$F_{1,3} = \frac{k(-Q_0)(+Q)}{x^2}$

The force on $q_3$ due to $q_2$ is

$F_{2,3} = \frac{k(-4Q_0)(+Q)}{1-x^2}$

The condition of equilibrium is

$F_{1,3} = F_{2,3}$

$\frac{k(-Q_0)(+Q)}{x^2}= \frac{k(-4Q_0)(+Q)}{1-x^2}$

$\frac{1}{x^2} = \frac{4}{(1-x)^2}$

$x = 0.331$ from $q_1$

To find the magnitude of $q_3$ we use $F_{1,2} = F_{1,3}$

$\frac{k(-Q_0)(4Q_0)}{1^2}= \frac{k(-Q_0)(Q)}{0.331^2}$

$Q = 0.43Q_0$

The magnitude of the third charge must be 0.43 the first charge $Q_0$

3 0

3 years ago

How does overproduction impact natural selection?

vova2212 [387]

Overproduction is a driving force in natural selection, as it can lead to adaptation and variations in a species. Darwin argued that all species overproduce, since they have more offspring than can realistically reach reproductive age, based on the resources available.

7 0

3 years ago

Read 2 more answers

Determine the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Eart

Temka [501]

The distance from the Earth's center to the point outside the Earth is 55800 Km

<h3>How to determine the distance from the surface of the Earth</h3>

Acceleration due to gravity of Earth = 9.8 m/s²
Acceleration due to gravity of the poin (g) = 1/60 × 9.8 = 0.163 m/s²
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Mass of the Earth (M) = 5.97×10²⁴ Kg
Distance from the surface of the Earth (r) =?

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by g

r² = GM / g

Take the square root of both sides

r = √(GM / g)

r = √[(6.67×10¯¹¹ × 5.97×10²⁴) / 0.163)]

r = 4.94×10⁷ m

Divide by 1000 to express in Km

r = 4.94×10⁷ / 1000

r = 4.94×10⁴ Km

<h3>How to determine the distance from the center of the Earth</h3>

Distance from the surface of the Earth (r) = 4.94×10⁴ Km
Radius of the Earth (R) = 6400 Km
Distance from the centre of the Earth =?

Distance from the centre of the Earth = R + r

Distance from the centre of the Earth = 6400 + 4.94×10⁴

Distance from the centre of the Earth = 55800 Km

Learn more about gravitational force:

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5 0

2 years ago

If a car increases its velocity from 1m/s to 3.6km/hr in 5 seconds what’s its acceleration

EleoNora [17]

Answer:

applying 1st eq of motion vf=vi+at here we have to find a=vf-vi/t , a= 1-1/5 , a=0/5 then we got a=0 here(vf value 3.6km/h is converted in standard unit 3.6×1000/3600 so we get vf=1m/s²

7 0

3 years ago

What is a lot like gases but the atoms are different because they are made up of free electrons and ions of elements

34kurt

plasmas are a lot like gases

hope this helps.

7 0

3 years ago