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storchak [24]
3 years ago
8

A 0.1 m by 0.1 m sheet of cardboard is placed in a uniform electric field of 10 N/C. At first, the plane of the sheet is oriente

d perpendicular to the electric field vector so that the electric flux through the sheet is 0.01 N-m2/C. By what angle do you need to rotate the sheet to reduce the electric flux by 1/2?
Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

The angle is 89°.

Explanation:

Given that,

Electric field = 10 N/C

Electric flux = 0.01 N-m²/C

Area A=\pi\times(0.1)^2

We need to calculate the angle

Using formula of electric flux

\phi=EA\cos\theta

\cos\theta=\dfrac{\phi}{EA}

Where, E = electric field

\phi = electric flux

A = area

Put the value into the formula

\cos\theta=\dfrac{\dfrac{0.01}{2}}{10\times\pi\times(0.1)^2}

\theta=\cos^{-1}(0.01592)

\theta=89.0^{\circ}

Hence, The angle is 89°.

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Ksp for agbr is 5x10-13. what is the maximum concentration of silver ion that you can have in a 0.1 m solution of nabr?
liberstina [14]

Answer : The maximum concentration of silver ion is 5\times 10^{-12}m

Solution : Given,

K_{sp} for AgBr = 5\times 10^{-13}

Concentration of NaBr solution = 0.1 m

The equilibrium reaction for NaBr solution is,

NaBr(aq)\rightleftharpoons Na^++Br^-

The concentration of NaBr solution is 0.1 m that means,

[Na^+]=[Br^-]=0.1m

The equilibrium reaction for AgBr is,

                          AgBr\rightleftharpoons Ag^++Br^-

At equilibrium                     s       s

The expression for solubility product constant for AgBr is,

K_{sp}=[Ag^+][Br^-]

The concentration of Ag^+ = s

The concentration of Br^- = 0.1 + s

Now put all the given values in K_{sp} expression, we get

5\times 10^{-13}=(s)(0.1+s)

By rearranging the terms, we get the value of 's'

s=5\times 10^{-12}m

Therefore, the maximum concentration of silver ion is 5\times 10^{-12}m.

4 0
2 years ago
Read 2 more answers
For a Physics course containing 10 students, the maximum point total for the quarter was 200. The point totals for the 10 studen
Talja [164]

Answer:

130.5

Explanation:

According to the stemplot attached (Which I think it is, and if not, then you only need to replace the procedure with your data and you should be fine), you need to calculate first the points of all ten students. In that plot, we can easily calculate the points.

The first number in the colum represents the centen of the point, while the numbers of the second column, represents the units of that centen, for example if you see:

16 | 8 5 6

This means that the point for the students are 168, 165 and 166. Three students, three points.

If you watch the stemplot, the points for the students are:

116, 118, 121, 124, 128, 133, 137, 142, 146 and 179.

The median can be calculated using the mean between the two values in the middle of the sequence.

In this case, half of ten is 5, so, the numbers from the middle in this sequence are 128 and 133, therefore:

Median = 128 + 133 / 2 = 130.5

5 0
3 years ago
A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest.
Tomtit [17]

Answer:

v = 6.79 m/s

Explanation:

It is given that,

Mass of a train car, m₁ = 11000 kg

Speed of train car, u₁ = 21 m/s

Mass of other train car, m₂ = 23000 kg

Initially, the other train car is at rest, u₂ = 0

It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,

m_1u_1+m_2u_2=(m_1+m_2)V

V is the common velocity after the collisions

V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s

So, the two car train will move with a common velocity of 6.79 m/s.

6 0
2 years ago
Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp
kotegsom [21]

Answer:

m2  = 83.3 g

Explanation:

by conservation of momentum principle we have

m_1v_{i1} + m_2v_{i2} = m_2v_{f2}

as both sphere has same speed so v_{i2} = v_{i1}

m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}

from conservation of kinetic energy principle we have

\frac{1}{2}m_1v^{2}_{i1} + \frac{1}{2}m_2v^{2}_{i2} = \frac{1}{2}m_2v^{2}_f2

v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}

v_{f1} =  v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}

\frac{v_{f1}}{v_{i2}} =\sqrt {\frac{(m_1+m_2)}{m_2}

substituting this value in above equation to get m2 value

m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}

solving for m2 we  get

m2 = \frac{m_1}{3}

m_1 = 250 g

      =\frac{250}{3}

  m2  = 83.3 g

7 0
3 years ago
9. A cube has a mass of 40g, a volume of 8cm' and a length of 2. What<br>is the density?​
Goshia [24]

Answer:

Density=mass/volume 40 divided by 8 =5

8 0
3 years ago
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