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hoa [83]
2 years ago
6

How much heat is removed from 60 grams of steam at 100 °C to change it to 60 grams

Physics
1 answer:
Harrizon [31]2 years ago
4 0

Answer:

45200J

Explanation:

Given parameters:

Heat of vaporization of water  = 2260J/g

Mass of steam = 20g

Temperature = 100°C

Unknown:

Energy released during the condensation  = ?

Solution:

This change is a phase change and there is no change in temperature

To find the amount of heat released;

         H  = mL

m is the mass

L is the latent heat of vaporization

Insert the parameters and solve;

         H  = 20g x 2260J/g

          H = 45200J

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Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

K =1/4 \pi \epsilon_0 = 8.99 \times 10^9 N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

E_1 + E_3 = 0

E =E_2 + E_4

E = 2 E_2

[E_2 =\frac{2\times k \times q}{r^2}

[r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m]

plugging all value

E = 2 E_2

E = 2 E_2 =\frac{2\times k \times q}{r^2}

E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}

E = 440816.32 N/C

3 0
3 years ago
Read 2 more answers
Help me plzzz I need answers
soldier1979 [14.2K]

Answer:

i think it is B

Explanation:

4 0
3 years ago
Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd
Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0
3 years ago
I need help with this
fredd [130]
We have here what is known as parallel combination of resistors.

Using the relation:

\frac{1}{ r_{eff} } = \frac{1}{ r_{1} } + \frac{1}{ r_{2} } + \frac{1}{ r_{3} }.. . + \frac{1}{ r_{n} } \\
And then we can turn take the inverse to get the effective resistance.

Where r is the magnitude of the resistance offered by each resistor.

In this case we have,
(every term has an mho in the end)
\frac{1}{10000} + \frac{1}{2000} + \frac{1}{1000} \\ \\ = \frac{1}{1000} ( \frac{1}{10} + \frac{1}{2} + \frac{1}{1} ) \\ \\ = \frac{1}{1000} ( \frac{31}{20}) \\ \\ = \frac{31}{20000}

To ger effective resistance take the inverse:
we get,
\frac{20000}{31} \: ohm \\ = 645 .16 \: ohm

The potential difference is of 9V.

So the current flowing using ohm's law,

V = IR

will be, 0.0139 Amperes.
7 0
3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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