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Deffense [45]
2 years ago
13

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Physics
1 answer:
Bad White [126]2 years ago
6 0

Answer:

\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}

\displaystyle \theta =68^o

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

\displaystyle F=\frac{KQ_1Q_2}{d^2}

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}

\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l

The force F3 is

\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle

F2 is pointed upwards and its horizontal component is zero

\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle

\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

Now we compute the total force

\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}

\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle

\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle

\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}

Now we compute the magnitude

\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}

The direction of the total force is given by

\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558

\boxed{\displaystyle \theta =68^o}

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Object A is 71 degrees and object B is 75 degrees .

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How long( in hours, will it take for 500 000 C of charge to flow through a diode if it requires
Alina [70]

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

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Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs  = 1,000,000/3600

1,000,000secs = 277.78 hours

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6 0
3 years ago
A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

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3 years ago
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3 years ago
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Conservation equations are powerful equations, because of how versatile and flexible they are to different physical situations.
Temka [501]

Answer:

b) false

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