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s2008m [1.1K]
3 years ago
13

Sound is a mechanical wave; therefore, it __________.

Physics
2 answers:
avanturin [10]3 years ago
8 0
A mechanical wave such as sound needs a medium to travel through. This is why they say that in space, no one can hear you scream (because space is a vacuum and no medium exists). On the contrary, light can exist in wave or particle form. This is why we have the ability to see beyond the realms of the Earth.
miss Akunina [59]3 years ago
6 0

Answer:

The answer is A. Must travel trought a medium.

Explanation:

The transmission of sound depends on the existence of a material medium (solid, liquid, air particles) where the vibration of the molecules are perceived as a sound wave, which are converted into mechanical waves in the human ear, to be later perceived by the brain.

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What does a battery powered heater warms up from ?
Westkost [7]
Battery based heaters use electric resistance heating, which uses a lot of current (electricity) to create the heat.
8 0
3 years ago
Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
kirza4 [7]

Answer:

The value is U  = 0.06 \  J

Explanation:

From the question we are told that

The value of charge on each three point charge is

q_1 = q_2 = q_3 =q=  1.05 \mu C  =  1.05 *10^{-6} \  C

The length of the sides of the equilateral triangle is r  =  0.500 \

Generally the total potential energy is mathematically represented as

U  = k *  [ \frac{q_1 *  q_2}{r}  +  \frac{q_2 *  q_3}{r}   + \frac{q_3 *  q_1}{r} ]

=> U  = k * 3 * \frac{q^2}{r}

Here k is coulomb constant with value k = 9*10^{9}\  kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=>    U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }

U  = 0.06 \  J

6 0
3 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
3 years ago
D
GenaCL600 [577]

Answer:

I don't know about these problems

Explanation:

I don't know about these problems at all.

7 0
3 years ago
180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperat
Zolol [24]

Answer : The final temperature of the mixture is 91.9^oC

Explanation :

First we have to calculate the mass of water.

Mass = Density × Volume

Density of water = 1.00 g/mL

Mass = 1.00 g/mL × 180 cm³ = 180 g

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of hot water (liquid) = 4.18J/g^oC

c_2 = specific heat of ice (solid)= 2.10J/g^oC

m_1 = mass of hot water = 180 g

m_2 = mass of ice = 20 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of hot water = 97^oC

T_2 = initial temperature of ice = 0^oC

Now put all the given values in the above formula, we get

(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC

T_f=91.9^oC

Therefore, the final temperature of the mixture is 91.9^oC

8 0
3 years ago
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