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Yakvenalex [24]
4 years ago
15

A test charge of +3 µC is at a point P where the electric field due to the other charges is directed to the right and has a magn

itude of 4 106 N/C. If the test charge is replaced with a charge of -3 µC, what happens to the electric field at P?
Physics
1 answer:
umka2103 [35]4 years ago
5 0

Answer:

Electric field will remain same.

Explanation:

Given that

At initial condition ,charge q = +3μ C

Electric field E = 4106 N/c

As we know that

Electric field due to charge q

E=\dfrac{kq}{r^2}  

Now when charge is replaced by new charge q'= -3μ C

From the expression of electric field we can say that electric field will remain same from same quantity of electric charge.

So we can say that electric field will remain same.

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Suppose that F3 = 310 N Determine the magnitude of the resultant force F′=F2+F3.
rewona [7]

Answer:

F' = 169.45N

This is a vector addition involving two vectors. In order to do this correctly, we need to resolve each of those forces into their vertical and horizon components and sum them up accordingly (all vertical components summed together and all horizontal components summed together). Then the magnitude of the summation is found by taking the square root of the sun of the squares of the summations along the vertical and the horizontal.

Explanation:

See the attachment below for the full solution to the problem.

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6 0
3 years ago
In the generation of most anions, the energy change (kj/mol that _______ an electron is ________.
g100num [7]
Released                       taken in
<span>Anion are associated with exothermic process.</span>

7 0
3 years ago
The diagram is a real world example of the first and second laws of thermodynamics. It shows how thermal energy is used to gener
Dmitriy789 [7]

Answer:

First law: kinetic energy is used to turn an electric generator

Second law: some thermal energy is lost to the environment as it travels through the system

Explanation:

The first law of thermodynamics is known as the law of conservation of energy. It states that energy can neither be created nor destroyed but can only be transferred or changed from one form to another. When thermal energy is used to generate electricity, the kinetic energy of the steam is used to turn the electric generator (thereby producing electrical energy).

The second law of thermodynamics states that energy transfer or transformation leads to an increase in entropy resulting in the loss of energy. This law also states that as energy is transferred or transformed, some is lost in a form that is unusable. When thermal energy is used to generate electricity, some of the thermal energy is lost to the environment as it travels through the system.

7 0
3 years ago
Read 2 more answers
1. A wave with a frequency of 10 Hz has a wavelength of 3 meters. At what
quester [9]

Answer:

30 m\ s Ans .....

Explanation:

Data:

f = 10 Hz

w = 3 m

v = ?

Formula:

v = fw

Solution:

v = ( 10)(3)

v = 30 m\ s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

3 0
3 years ago
A 3-ft3 rigid tank initially contains saturated water vapor at 300°F. The tank is connected by a valve to a supply line that car
zhannawk [14.2K]

Explanation:

Given\\V=3ft^3\\T_{1} =300^oF\\T_{L} =400^oF\\P_{L} =200psia\\V_{f} =0.5V

We assume kinetic and potential energy changes are negligible and there is no work interactions.

a) Taking tank as a system, The energy balance can be define as

E_{in}- E_{out}= E_{sys}

m_{i} h_{L} -Q_{out}=m_{2}  u_{2}-m_{1}  u_{1}

The mass balance could be written as

m_{in}- m_{out}= m_{sys} \\m_{i}= m_{2}- m_{1}

The final pressure in the tank could be defined as following

P_{2} =P_{sat300^oF}

from standard steam table we know at

T_{2}= T_{1}=300^oF\\ P_{2}=67.028psia

b)

From steam table at

T_{1} =300^oF\\v_{1}= v_{g}=6.4663ft^3/lbm\\v_{1}=  v_{g}=1099.8Btu/lbm\\ v_{f} =0.01745ft^3/lbm\\v_{f} =269.51Btu/lbm\\

at\\P_{L} =200psia\\T_{L}=400^oF\\h_{L}=1210.9Btu/lbm

initial mass in the tank could be define as

m_{1}=\frac{V}{v_{1} }  \\m_{1} =\frac{3}{6.4663} =0.464lbm\\

Final mass in the tank could be define as

m_{2}=\frac{V_{f} }{v_{f} }+\frac{V_{g} }{v_{g} }  \\ m_{2} =\frac{1.5}{0.01745} +\frac{1.5}{6.4663} =86.2lbm

The amount of steam that has entered the tank

m_{i}=m_{2}-  m_{1}\\ m_{i}=86.2-0.464=85.74lbm

c)

The internal energy in final state could be defined as following

U_{2}=m_{f}  u_{f}+ m_{g} u_{g}\\ U_{2} =85.96*269.51+0.232*1099.8=23422Btu\\

The heat transfer could be defined as following

Q_{out}=m_{i}  h_{L}+m_{1}  u_{1}-m_{2}  u_{2}\\ Q_{out}=85.74*1210.9+0.464*1099.8-23422=80910Btu

6 0
3 years ago
Read 2 more answers
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