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3 years ago

Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:

Physics

1 answer:

Alexxx [7]3 years ago

8 0

Answer:

1).atoms (3). mixture. (5). Element

2). particles (4). molecules (6). suspension

Explanation:

(7). Homogeneous (8). Heterogeneous

(9). compound (10). solutions

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what is the position of centre of curvature for concave and convex mirror show with the help of diagram if you can

Anon25 [30]

it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram

8 0

3 years ago

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6

lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

The amplitude of induced emf by this signal between the ends of the receiving antenna is

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

Here,

I = Current

$\mu_0$ = Permeability at free space

c = Light speed

d = Distance

Replacing,

$\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}$

$\epsilon_{max} = 0.05434V$

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0

3 years ago

A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of

Gekata [30.6K]

Answer:

a) $P=2450\ Pa$

b) $\delta h=23.162\ cm$

Explanation:

Given:

height of water in one arm of the u-tube, $h_w=25\ cm=0.25\ m$

Gauge pressure at the water-mercury interface,:

$P=\rho_w.g.h_w$

we've the density of the water $=1000\ kg.m^{-3}$

$P=1000\times 9.8\times 0.25$

$P=2450\ Pa$

Now the same pressure is balanced by the mercury column in the other arm of the tube:

$\rho_w.g.h_w=\rho_m.g.h_m$

$1000\times 9.8\times 0.25=13600\times 9.8\times h_m$

$h_m=0.01838\ m=1.838\ cm$

<u>Now the difference in the column is :</u>

$\delta h=h_w-h_m$

$\delta h=25-1.838$

$\delta h=23.162\ cm$

7 0

2 years ago

Find the final velocity of a 40 kg skateboarder traveling at an initial velocity of 10 m/s that moves up a hill from a height of

slavikrds [6]

Answer:

vf = 0

Explanation:

Since the initial height hi = 0, we can rewrite the energy equation as

vf^2 = vi^2 - 2ghf = (10 m/s)^2 - 2(10 m/s^2)(5 m) = 0

Therefore, his final velocity vf is

vf = 0

3 0

3 years ago

PLEASE SHOW STEPS!

SOVA2 [1]

Answer:

660 J/kg/°C

Explanation:

Heat lost by metal = heat gained by water

-m₁C₁ΔT₁ = m₂C₂ΔT₂

-(0.45 kg) C₁ (21°C − 80°C) = (0.70 kg) (4200 J/kg/°C) (21°C − 15°C)

C₁ = 660 J/kg/°C

8 0

3 years ago