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3 years ago

Hi can you please help me, im really stuck

1 answer:

3 0

I have the answer for A. Since there is blockage in the ear canal, some sound waves may not be able to get through or travel as quickly so you would have trouble hearing

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Siruis, the brightest star in the night sky, has a luminosity of 22. This means that Sirius: A, B, C, D QUESTION

suter [353]

Answer:

i think C

Explanation:

4 0

2 years ago

Which diseases are most likely to be treated with antibiotics?

Vitek1552 [10]

Answer:

c) those caused by parasites

4 0

2 years ago

The distance between two charged objects is doubled. What happens to the electrostatic force between the two?a)It will double.b)

zzz [600]

Answer:

d) It will be cut to a fourth of the original force.

Explanation:

The magnitude of the electrostatic force between the charged objects is

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, the initial distance is doubled, so

r' = 2r

Therefore, the new electrostatic force will be

$F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F$

So, the force will be cut to 1/4 of the original value.

5 0

3 years ago

A stream moving with a speed of 7.1 m/s reaches a point where the cross-sectional area of the stream decreases to one half of th

lana66690 [7]

Answer:

14.2 m/s

Explanation:

Given data:

Speed of the stream, v₁ = 7.1 m/s

let the cross section area at initial point be A₁

now area at the second point, A₂ = (1/2)A₁ = 0.5A₁

now, from the continuity equation, we have

A₁v₁ = A₂v₂

where, v₂ is the velocity at the narrowed portion

thus, on substituting the values, we get

A₁ × 7.1 = 0.5A₁ × v₂

v₂ = 14.2 m/s

8 0

3 years ago

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

strojnjashka [21]

Answer:

$1.475\times 10^{-13}\ C/m^3$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area

h = Altitude = 600 m

Electric flux through the top would be

$-110A$ (negative as the electric field is going into the volume)

At the bottom

$120A$

Total flux through the volume

$\phi=120-110\\\Rightarrow \phi=10A$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0$

Charge per volume is given by

$\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3$

The volume charge density is $1.475\times 10^{-13}\ C/m^3$

7 0

3 years ago