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3 years ago

I WILL MARK YOU THE BRAINLIEST NO LINKS AND IF YOU DONT UNDERSTAND THE QUESTION DONT ANSWER

Physics

1 answer:

elena-s [515]3 years ago

6 0

Answer:

Perform a simple test of the material in the pan to assess whether it is real gold. Raw gold appears brassy yellow and bright. If you think it is gold, place your hand between it and the sun to create shade over the gold. If it still appears bright in the pan, chances are that it is real gold.

Explanation:

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A magnetic dipole <img src="https://tex.z-dn.net/?f=%5Cvec%7Bm%7D" id="TexFormula1" title="\vec{m}" alt="\vec{m}" align="absmidd

ser-zykov [4K]

We are asked to know the reason why magnetic dipole can be written as U = -m *B. The reason for this was that the negative sign is just showing or indicating the direction of the force. If it if negative, it means that the work done is opposite to the force field.

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3 years ago

The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b

ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})$

where;

Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²

k is the dielectric constant = 2.3

b is the outer radius of the conductor = 3 mm

a is the inner radius of the conductor = 0.8 mm

λ is the linear charge density = 18 x 10⁶ V/m

Substitute in these values in the above equation;

$V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) = \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V$

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V

8 0

3 years ago

Why is a concave mirror is used a reflector in a torch light?

tatiyna

Answer:

diverging light rays of the bulb are collected by the reflector.

Explanation:

4 0

3 years ago

Read 2 more answers

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$