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3 years ago

Full moon is located______

Physics

2 answers:

nekit [7.7K]3 years ago

7 0

Answer:

Farthest from the Sun

Tasya [4]3 years ago

6 0

Answer:

Im going to take an educated guess and say its B.

Explanation:

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You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-

Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

$E_P=\int dE \cos$

$E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}$

$\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ$

$\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}$

If we put an electron on point P, then force on point e is :

$\vec{F}=-|e|\vec{E_P}$

$F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}$

If r >> x , then $\frac{x^2}{r^2} \approx 0$

Then, $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0

3 years ago

Question 2 of 10

emmainna [20.7K]

Answer:

Explanation:

I just did it on apex

8 0

3 years ago

QUESTION 7

harina [27]

The term used is " WORK".

I'm really sorry it's not one of the choices.

8 0

3 years ago

Read 2 more answers

A bicycle travels 141 m along a circular track of radius 30 m. What is the angular displacement in radians of the bicycle from i

Veseljchak [2.6K]

Answer: 4.7rad

Explanation:

Angular displacement =s/r

Where s=distance traveled

r=radius

Angular displacement =141m/30m

Angular displacement =4.7rad.

5 0

3 years ago

A wire, of length L = 4.1 mm, on a circuit board carries a current of I = 1.96 μA in the j direction. A nearby circuit element g

tamaranim1 [39]

Answer:

The magnitude of the magnetic field B is 5.921 T.

Explanation:

Given that,

Length = 4.1 mm

$B_{x}=4.9\ G$

$B_{y}=2.3\ G$

$B_{z}=2.4\ G$

Current $I = 1.96\ mu A$

We need to calculate the magnetic field

Using formula of magnetic field

$B=\sqrt{B_{x}^2+B_{y}^2+B_{z}^2}$

Put the value into the formula

$B=\sqrt{(4.9)^2+(2.3)^2+(2.4)^2}$

$B=5.921\ T$

Hence, The magnitude of the magnetic field B is 5.921 T.

6 0

3 years ago