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iren2701 [21]
3 years ago
9

Sentence A: At the same time, teachers will benefit from teaching fewer students per semester and gaining more one-on-one time w

ith students Sentence B: For all of these reasons, a change from the traditional seven-period class schedule to more compact, four-class block schedule will be beneficial for both students and teachers. Sentence C: Students will enjoy less homework each night, better focus due to fewer subjects, and more variety, with new classes each semester. Sentence D: When it comes to exchanging the seven- period day to a four-period day, less is definitely more. Which arrangement of these sentences would be best for a conclusion paragraph?
A. B, C, A, D
B. B, D, A, C
C. D, A, C, B
D. C, A, B, D
plz help me
Physics
2 answers:
frez [133]3 years ago
7 0

I think its c

d a c b

I hope this is helpful

QveST [7]3 years ago
4 0

Answer:

Its A.  B, C, A, D

Explanation:

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A student performed the following steps to find the solution to the equation x^2-2x-15=0 Where did the student go wrong? STEP 1:
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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

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r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

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THE PROBLEM!
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