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Snowcat [4.5K]
3 years ago
14

In the heat equation what does Q represent

Physics
2 answers:
Burka [1]3 years ago
6 0
Q stands for heat energy (Joules) hope this helps!
svlad2 [7]3 years ago
3 0

A. heat required to raise the temperature

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A neutral atom of uranium (U) has 146 neutrons and a mass number of 238. How many protons does it have?
BARSIC [14]

Answer:

92 protons

Explanation:

The mass number is 238 , so the nucleus has  <u>238  particles</u> in total, including <u>146 neutrons</u>.  So to calculate the number of neutrons we have to subtract:  238  −  146  =  92

4 0
3 years ago
Read 2 more answers
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).

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
3 years ago
What is the amount of work done when a force of 10N moves a 20kg mass of 8 meters?
Nataly_w [17]

Yo sup??

we should know that

work done=force*displacement

W=F.s

F=10 N

s=8 m

therefore

W=10*8

=80 N

Hope this helps.

3 0
2 years ago
3. During a tug-of-war, Team A pulls with a
Neko [114]

Answer:

8000 - 5000 =3000

Explanation:

8 0
2 years ago
Read 2 more answers
As a consultant to the soft-drink industry, Dr. J is given the task of conducting the ultimate Pepsi taste test. This is Dr. J's
dem82 [27]

Answer:

A) How much work does the Pepsi do on the bullet = 0.0625J

B) At what velocity does the Pepsi hit the floor = 7.67m/s

Explanation:

  • Given mass of bullet = 5g
  • initial velocity = 500 m/sec
  • final velocity = 5 m/sec

From work done = Force X Distance

  • WD = Mad
  • a = v - u /t

Workdone ; m(v - u )/t

A) The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi:

  • but The work done by the container wall on the bullet is equal to the change in kinetic energy on either side of the wall: W = change in kinetic energy = 1/2 mv²

= 1/2 X 0.005 x 500² - 1/2 X 0.055 x 5²

= 624.94J

  • Therefore ; The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi = W = change in KE = 0.0625 - 0 = 0.0625J

B) At what velocity does the Pepsi hit the floor?

  • From conservation of energy principle; PE = KE
  • mgh = 1/2mv²
  • 0.98 X 3 = 1/2v²
  • velocity = 7.67m/s

3 0
3 years ago
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