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3 years ago

Need answer ASAP!!!

Physics

1 answer:

Alina [70]3 years ago

5 0

Hello!

Use <u>ohm law:</u>

I = V / R

Replacing:

I = 100 V / 20 O

I = 5 A

The current is <u>5 amperes.</u>

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

Bad White [126]

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

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

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

6 0

3 years ago

Shelli is a student describing the resonance of a metal beam. She demonstrates that the sound dims when the beam is vibrated at

ArbitrLikvidat [17]

Shelli will most likely say that the sound produced by the resonance of a metal is affected by the frequency at which the metal is vibrated. There is a certain frequency that will create a dim sound. If the frequency is changed to the right setting, the sound will intensify.

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4 years ago

A nichrome wire has a resistance of 5

kari74 [83]

Answer:

R = 6.67 Ohm's

Explanation:

Resistance is a property of a material that measures the opposition to the flow of current through the material. It is measured in Ohm's.

R = (ρl) ÷ A

where; R is the resistance of the material, A is its cross sectional area, l is its length and ρ is its resistivity.

For the first nichrome wire, resistance is 5 Ohm's.

i.e R = (ρl) ÷ A = 5 Ohm's

For the second nichrome wire, length = 4l, and area of cross section = 3A. The resistance of the second wire can be determined by;

R = (4ρl) ÷ 3A

= $\frac{4}{3}$ {(ρl) ÷ A}

= $\frac{4}{3}$ × 5

R = 6.67 Ohm's

Resistance of the second nichrome wire is 6.67 Ohm's.

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