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3 years ago

The enthalpy of fusion of ice is 330,000 J/kg. If I have a 1 kg sample, how much energy is needed to melt all of it?

Physics

2 answers:

EleoNora [17]3 years ago

8 0

Answer:

C. 330,000 J

Explanation:

Latent heat = mass × specific latent heat

q = mL

Given m = 1 kg and L = 330,000 J/kg:

q = (1 kg) (330,000 J/kg)

q = 330,000 J

Anestetic [448]3 years ago

6 0

Answer:

C. 330,000 J

Explanation:

The enthalpy of fusion of ice is 330,000 J/kg. If someone has a 1 kg sample, they will need 330,000 J energy in order to melt all of it.

(1 kg) (330,000 J/kg)

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Typical Pressurized Water Reactors can produce 1100 to 1500

lozanna [386]

Answer: about 1,100,000,000 to 1,500,000,000 Joules/second

Explanation:

1 MW (megawatt) = 1,000,000.00 J/s (joules per second)

1100(1,000,000) = 1,100,000,000

1500(1,000,000) = 1,500,000,000

3 0

3 years ago

Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor

fenix001 [56]

Answer:

a) $F_{r}= -583.72MN i + 183.47MN j + 6.05GN k$

b) $E=3.04 \frac{GN}{C}$

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

$r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}$

which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

$F=k_{e}\frac{q_{1}q_{2}}{r^{2}}$

so we substitute the values:

$F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}$

which yields:

$F_{AP}=418.14 MN$

Now we can find its components:

$F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i$

$F_{APx}=191.30 MNi$

$F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j$

$F_{APy}=191.30MN j$

$F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k$

$F_{APz}=318.83 MN k$

And we can now write them together for the first force, so we get:

$F_{AP}=(191.30i+191.30j+318.83k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}$

which yields:

$r_{BP}^{2}= 33 m^{2}$

Next, we can make use of the force formula:

$F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}$

which yields:

$F_{BP}=2.18 GN$

Now we can find its components:

$F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i$

$F_{BPx}=758.98 MNi$

$F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j$

$F_{BPy}=758.98MN j$

$F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k$

$F_{BPz}=1.897 GN k$

And we can now write them together for the second, so we get:

$F_{BP}=(758.98i + 758.98j + 1897k)MN$

We continue with the next force. The procedure is the same so we get:

$r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}$

which yields:

$r_{CP}^{2}= 30 m^{2}$

Next, we can make use of the force formula:

$F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}$

which yields:

$F_{CP}=4.20 GN$

Now we can find its components:

$F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i$

$F_{CPx}=-1.534 GNi$

$F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j$

$F_{CPy}=-766.81 MN j$

$F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k$

$F_{CPz}=3.83 GN k$

And we can now write them together for the third force, so we get:

$F_{CP}=(-1.534i - 0.76681j +3.83k)GN$

So in order to find the resultant force, we need to add the forces together:

$F_{r}=F_{AP}+F_{BP}+F_{CP}$

so we get:

$F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN$

So when adding the problem together we get that:

$F_{r}=(-0.583.72i + 0.18347j +6.05k)GN$

which is the answer to part a), now let's take a look at part b).

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

$F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}$

which yields:

$F_{r}=6.08 GN$

and now we take the formula for the electric field which is:

$E=\frac{F_{r}}{q}$

so we go ahead and substitute:

$E=\frac{6.08GN}{2C}$

$E=3.04\frac{GN}{C}$

7 0

4 years ago

PLeaSe SoMeBodY HeLp

Lerok [7]

1). Oört Cloud
2). Kuiper Belt
3). asteroids
4). meteorite

Note:
Your teacher doesn't need this information !
Your teacher needs for you to LEARN stuff.
You have the answers now, but until you memorize the information,
you haven't learned anything.

5 0

3 years ago

Reactive elements such as alkali metals and halogens are found in nature only as

bonufazy [111]

Compounds

Explanation:

Reactive elements such as alkali metals and halogens are found in nature only as compounds. Such elements are too unstable to remain as stable atoms, therefore they readily combine and form compounds.

Compounds are formed when two atoms combines together to share electrons.
They either lose, gain, or share electrons between themselves.
In the end, they end up becoming more stable.
This is the reason why atoms combine.
Unstable elements are very reactive especially alkali metals and halogens.
On their own, they are unstable and prefers to bond with other atoms in order to gain a measure of stability.
This is why they are found in combined state in nature.

Learn more:

Compound brainly.com/question/10585691

Noble gases brainly.com/question/1781595

#learnwithBrainly

4 0

3 years ago

The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one stud

motikmotik

<h2>The distance between students is 2.46 m</h2>

Explanation:

The force of attraction due to Newton's gravitation law is

F = $\frac{Gm_1m_2}{r^2}$