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Ronch [10]
3 years ago
14

The closer the particles are together, the ______ sound is able to move.

Physics
1 answer:
maksim [4K]3 years ago
4 0

Answer:

Faster

Explanation:

The closer the particles are together, the faster sound is able to move.

Sound travels the slowest through the air because gas particles are more spread apart than liquids or solids.

Sound will travel the fastest through solids because the particles in solids are densely packed together.

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1. Have you ever tried to undergo an X- ray to test on youur body? What can you see in the fiom after the examination? Why is th
liraira [26]

Answer:

1) λ < 2d,  2)  nfrared imaging technique, 3) each color there is a different index of refraction

Explanation:

We are going to answer the three questions

1) When x-rays pass through matter in order to be dispersed, their wavelength must be of the order of the length of separation in the atoms and molecules of the body, in solid bones this length is similar and they scatter and reflect the x-rays therefore they can be observed, the fat and the soft tissue have a much greater separation therefore the x-rays cannot be reflected and consequently it is not observable by this technique.

2) At airports they use the infrared imaging technique, where the image is taken for the infrared wavelength, which is the heat part of the electromagnetic spectrum; consequently, when the image is viewed, the hottest areas appear brighter and, since when a person has a virus, his temperature rises, his temperature rises, it is possible to observe people with a higher temperature.

3) when white light hits a prism it is refracted with the equation

            n₁ sin θ₁ = n₂ sin θ₂

where the incidence of refraction depends on the wavelength, therefore for each color there is a different index of refraction and consequently the light is separated in its different colors.

5 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
Read 2 more answers
Why does the cord of an electric heater not glow while the heating element does​
Strike441 [17]

Answer: Why does the cord of an electric heater not glow while the heating element does? The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminum is very low so it does not glow.

Explanation:

7 0
3 years ago
A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun
Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{100.0}{300}

t=0.33\ sec

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

s=0+\dfrac{1}{2}\times9.8\times(0.33)^2

s=0.53\ m

Hence, The distance is 0.53 m.

5 0
3 years ago
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
3 years ago
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