All categories

3 years ago

5.Noise pollution is harmful forRequired to answer. Single choice.Immersive Reader

Physics

1 answer:

son4ous [18]3 years ago

3 0

Answer:

D: All

Explanation:

In cats, noise pollution causes what is known as acoustic stress in them.

In birds, noise pollution delays their nesting because it makes their songs which they use to probe to be at a low frequency.

Noise pollution affects humans because it can lead to loss of hearing, sleep disturbance, stress, high blood pressure e.t.c

Therefore, it affects Cats, birds and humans.

So option D is correct.

You might be interested in

What is the difference between a wave and a medium?

olga nikolaevna [1]

Answer:

Mediums in which the speed of sound is different generally have differing acoustic impedances, so that, when a sound wave strikes an interface between

Explanation:The propagation of a wave through a medium will depend on the properties of the medium. For example, waves of different frequencies may travel

7 0

3 years ago

Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sp

Leokris [45]

Answer:

A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

V = k ∑ $q_{i} / r_{i}$

where $q_{i}$ and $r_{i}$ are the loads and the point distances.

A) We apply this equation to our case

V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)

V = 9 10¹ / 1.65 (1.10 - 3.60)

V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

r₂ = 3.30 -0.02 = 3.28 m

r₁ = 0.02 m

we calculate

V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)

V = 9 10¹ (55 - 1,098)

V = 4.85 10³ V

C) The potential on the surface of sphere B

r₂ = 0.02 m

r₁ = 3.3 -0.02 = 3.28 m

V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)

V = 9 10¹ (0.335 - 180)

V = 1.62 10⁴ V

6 0

3 years ago

A house is advertised as having 1560 square feet under roof. What is the area of this house in square meters?

strojnjashka [21]

Answer:

1 m = 3.28 ft

1 m^2 = 10.76 ft^2

1560 ft^2 / 10.76 ft^2 / m^2 = 145 m^2

7 0

3 years ago

A cannonball is fired vertically upwards at 100.0 m/s a) How long will it take to return to the cannon? b) what is it's maximum

V125BC [204]

Answer:
a) 20s
b) 500m

Explanation:
Given the initial velocity = 100 m/s, acceleration = -10m/s^2 (since it is moving up, acceleration is negative), and at the maximum height, the ball is not moving so final velocity = 0 m/s.

To find time, we apply the UARM formula:

v final = (a x t) + v initial

Replacing the values gives us:

0 = (-10 x t) + 100

-100 = -10t

t = 10s

It takes 10s for the the ball to reach its max height, but it must also go down so it takes 2 trips, once going up and then another one going down, both of which take the same time to occur

So 10s going up and another 10s going down:

10x2 = 20s

b) Now that we have v final = 0, v initial = 100, a = -10, t = 10s (10s because maximum displacement means the displacement from the ground to the max height) we can easily find the displacement by applying the second formula of UARM:

Δy = (1/2)(a)(t^2) + (v initial)(t)

Replacing the values gives us:

Δy = (1/2)(-10)(10^2) + (100)(10)

= (-5)(100) + 1000

= -500 + 1000

= 500 m

Hope this helps, brainliest would be appreciated :)

7 0

3 years ago

A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

3 years ago