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3 years ago

PLEASE HELP ME !!!!!

Physics

1 answer:

telo118 [61]3 years ago

4 0

Answer: radiant energy

Explanation:

The energy in electromagnetic waves is sometimes called radiant energy.

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A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

Therefore the heat lost by mercury is 0.09J

5 0

3 years ago

A vertical piston-cylinder device contains water and is being heated on top of a range. During the process, 63 Btu of heat is tr

jolli1 [7]

Answer:

The water would have a change in the energy of 50 BTU

Explanation:

Under the assumption that the piston-cylinder system is a closed system, then the conservation law of the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all the energy that goes into the system, $E_{out}$ is all the energy that leaves the system and, $E_{change}$ is the change of the energy of the system.
For this case the energy that is feeded to the system is heat: 63 BTU.
There is two ways that energy is leaving this system: 8 BTU of heat as losses through the side walls and, 5 BTU of work done by the vapor on the piston.
Then the change of energy of the system is: $E_{change}=E_{in}-E_{out}=63 BTU-(8BTU+5BTU)=50BTU$ .

6 0

3 years ago

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

8 0

3 years ago

Help me out please I need to find this out quick

photoshop1234 [79]

He got u too bad for the late reply to the question I have;)

6 0

3 years ago

Read 2 more answers

If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?

Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

$R=A=B$

So, using the law of cosines, we have:

$R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°$

Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°

Have a nice day!

4 0

3 years ago