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Elodia [21]
3 years ago
9

The masses of blocks A and B in the figure (Figure 1) are 20.0 kg and 10.0 kg, respectively. The blocks are initially at rest on

the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F⃗ is applied to the pulley.
Find the acceleration a⃗ A of block A when F is 124 N.
Physics
1 answer:
Anna35 [415]3 years ago
7 0
<span>You are given the masses of blocks A and B in the figure are 20.0 kg and 10.0 kg, respectively. The blocks are initially at rest on the floor and are connected by a massless string passing over a massless and frictionless pulley. An upward force F⃗ is applied to the pulley. The acceleration is 4.13 m/s^2.

</span>
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<span>6160 joules

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3 years ago
A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A
MakcuM [25]

Answer:

The magnitude and algebraic sign of q is 14\sqrt{2}\ \mu C

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}

The electric force at opposite corner

F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}

The net force is

F=\sqrt{F_{1}^2+F_{2}^2}

Put the value into the formula

F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}

The electric force at second corner

F_{2}=\dfrac{-kq^2}{2r^2}

The net force acting on either of the charges is zero.

So,  F=F'

\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}

\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}

q=14\sqrt{2}\ \mu C

Hence, The magnitude and algebraic sign of q is 14\sqrt{2}\ \mu C

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vekshin1

Answer:

7.37 kg

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Mass = 140N / 19m/s^2

= 7.37 kg

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