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3 years ago

A student has the following equipment - copper wire, nail made of iron,a battery, paperclips.

1 answer:

3 0

Taking the copper wire, he has to wind it around the nail made of iron. After which, he then connect both ends of the copper wire to the battery, so an electric charge travels through the wire. This is the basic electromagnet. Since a current is now flowing through the wire, a magnetic field is produced. Placing the electromagnet near the mixture of copper and iron, the magnet should attract the pieces of iron, as iron is more magnetic compared to copper. This is done over a period of time, so that only the copper pieces are left in the mixture.

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What type of radiation does not have to be observed above earths atmosphere? Visible light, X rays, Gamma rays, Ultraviolet radi

juin [17]

The correct answer is visible light.

Hope this helps.

6 0

2 years ago

Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom

Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

final velocity of Tarzan = v_f

law of conservation of energy

PE_i + KE_i = PE_f + KE_f

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i = \dfrac{1}{2}mv_f^2$

$v_f = \sqrt{2gh_i}$

$= \sqrt{2gL(1- cos\theta)}$

$= \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}$

= 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the = 3 m/s

$mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2$

$mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2$

$mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2$

$gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2$

$v_f = \sqrt{v_1^2+2gh_i}$

$v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}$

v_f= 11.29 m/s

3 0

3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$