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3 years ago

How does the intensity of a sound wave change if the distance from the

Physics

1 answer:

goblinko [34]3 years ago

3 0

Answer:

the answer is that it increases by a factor of 16

Explanation:

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What is the difference between AM radio waves and FM radio waves?

N76 [4]

The correct answer is B.

6 0

3 years ago

Read 2 more answers

True [87]

Answer:

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

Explanation:

Given:

mass of ice, $m_i=14.7\ g=0.0147\ kg$

mass of water, $m_w=324\ g=0.324\ kg$

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

$Q_i=Q_w$

$m_i.L=m_w.c_w.\Delta T$ ......................(1)

where:

$L=$ latent heat of fusion of ice $=333.55\ J.g^{-1}$

$c_w=$ specific heat of water $=4.186\ J.g^{-1}.^{\circ}C^{-1}$

$\Delta T=$ change in temperature

Putting values in eq. (1):

$14.7 \times 333.55=324\times 4.186\times \Delta T$

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

8 0

3 years ago

The electric field in a particular thundercloud is 3.8 x 105 N/C. What is the acceleration (in m/s2) of an electron in this fiel

Svetradugi [14.3K]

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

8 0

3 years ago

Please help! Will give brainly, 50 points!! I'm stuck with this question and I don't get it!!

IrinaVladis [17]

Answer:

3.52176 x 10^-10 N

Explanation:

Fg = 3.52176 x 10^-10 Newton

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