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3 years ago

Hey plz help mesuggest a situations where you might need to use ear defenders

Physics

1 answer:

xenn [34]3 years ago

4 0

You would need to ear defenders (I'm assuming these are ear protectors for use with loud sounds) while firing at a gun range. You could use it for loud construction areas.

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What flows in electricity?<br> A. Electrons<br> B. Protons<br> C. Neutrons

nordsb [41]

Answer:

Explanation:

Electrons

6 0

3 years ago

Two objects separated by a distance r are each carrying a charge q The magnitude of the force exerted on the second object by th

Sindrei [870]

Answer:F=4F

Explanation: Columbs law states that The force between the two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them

Force between the two charges is given by

F=K*q1*q2/r^2

if one charge become 4 times, new force is,

F=4(K*q1*q2)/r^2

F=4F

Where q1 and q2 are the point charges

r is the distance between the two charges

K is a constant of proportion called electrostatic force

4 0

3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

4 years ago

When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a

Nana76 [90]

Answer:

Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

Фi = ∫ E .dA = $q_{int}$ / ε₀

the Gaussian surface is a sphere so its area is

A = 4 π r²

the charge inside is

q_{int} = Q

we substitute

E 4π r² = Q /ε₀

E = 1 / 4πε₀ Q / r²

To calculate the flow on the two surfaces

* Sphere

Ф = E A

Ф = 1 / 4πε₀ Q / r² (4π r²)

Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

r = d / 2

We look for the diagonal with the Pythagorean theorem

d² = L² + L² = 2 L²

d = √2 L

we substitute

r = √2 / 2 L

r = L / √2

L = √2 r

now we can calculate the area of the cube that has 6 faces

A = 6 L²

A = 6 (√2 r)²

A = 12 r²

the flow is

Ф = E A

Ф = 1 / 4πε₀ Q/r² (12r²)

Ф_cubo = 3 /πε₀ Q

the relationship of these two flows is

Ф_cube /Ф_sphere = 3 /π

8 0

4 years ago

What causes diffferent air density?

ololo11 [35]

Answer:

The temperature and gravity both affects the density

5 0

3 years ago