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2 years ago

Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.

Physics

2 answers:

m_a_m_a [10]2 years ago

3 0

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, $q_1=6\ \mu C=6\times 10^{-6}\ C$

Charge 2, $q_2=2\ \mu C=2\times 10^{-6}\ C$

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

$F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N$

So, the force applied between charges is 10.788 N.

Margaret [11]2 years ago

3 0

Answer:

10.8 N

Explanation:

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Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.

salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_{1} = u_{g}$ = 2553.6 kJ/kg

$v_{1} = v_{g} = 0.4625 m^{3}/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_{2} = v_{g} = 0.4625 m^{3}/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_{2} = v_{g} = 0.4625 m^{3}/kg$ and temperature $T_{2} = 360^{o}C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})$

$u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$\frac{Q}{m} = \Delta u$

$\frac{Q}{m} = u_{2} - u_{1}$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0

1 year ago

g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

$R = \frac{u^{2}Sin2\theta }{g}$

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

$R = \frac{u^{2}Sin60 }{9.8}$

R = 0.088u^2

If the angle of projection is 45 degree.

$R = \frac{u^{2}Sin90 }{g}$

R = u^2 / g

5 0

2 years ago

A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to 22.5N.

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that the force tending to lift the load off the ground (vertical force) is equal to 22.5N.

Have a nice day!

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Katie rolls a toy car off the end of a table.

hoa [83]

Answer: C.

Explanation:

i choose B and got it wrong it told me the answer after on edge.

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1 year ago

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sashaice [31]

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1 year ago