Define reflection of sound?​

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Answer 1
Answer: The reflection of sound is the movement of sound waves bouncing off of a surface and back into another direction, hope this helped :)

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Two charged particles are located on the x axis. The first is a charge +Q at x = −a. The second is an unknown charge located at x = +3a. The net electric field these charges produce at the origin has a magnitude of 2keQ/a2 . Explain how many values are possible for the unknown charge and find the possible values.

Answers

Answer:

-9Q

Explanation:

Electric field at origin is:

E=(2keQ)/(a^2)

Electric field due to first charge at origin would be:

E_1=(keQ)/(a^2)

Electric field due to second charge would be:

E_2=E-E_1\nE_2=(2keQ)/(a^2)-(keQ)/(a^2) = (keQ)/(a^2)

If the second charge is Q', then E_2 should be:

E_2=(keQ')/((3a)^2)=(keQ')/(9a^2)

compare the above two values to find the possible values of Q':

(|Q'|)/(9)=Q\n |Q'|=9Q

The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.

Thus, Q' = -9Q

One value is possible as the location of the second charge is given to be on the positive x-axis.

Final answer:

The possible values for the unknown charge are 1/9 of the magnitude of the known charge.

Explanation:

To find the possible values for the unknown charge, we need to use the principle of superposition. The net electric field at the origin is given by the sum of the electric fields due to each charge. We know that the magnitude of the net electric field is 2keQ/a^2, so we can set up the equation:

2keQ/a^2 = keQ/(-a)^2 - keq/(3a)^2

By solving this equation, we can find the possible values for the unknown charge. Simplifying the equation, we get:

2 = 1 - 1/9

1/9 = 1

After solving the equation, we find that the possible value for the unknown charge is 1/9 of the magnitude of the known charge.

Learn more about Electric field here:

brainly.com/question/8971780

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determine exactly where to place a cart on the track so that it rolls down the track, flies through the air, and lands precisely at 1) the green line, 2) the red line, and 3) the blue line, on the first try.

Answers

Answer: i think you should place it on the red line

Explanation:

hope this helps

and need brainliest

If the population of penguins increased, then this would have a direct effect on the populations of?

Answers

Answer:

Globel warming

Explanation:

hope this helpex

"Which gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string?"

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Answer:

the second derivative of y with respect to time gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string

Explanation:

This is because the transverse wave movement of particles take place in direction 90° to direction of movement of the wave (x) itself, so second derivative of y with respect to time (t)is what will be required

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change the period to 2.07 s?

Answers

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1  (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

Why must the Ojibwe people pay close attention to the seasons? a.) they must be ready to move to a new place where they can hunt

b.) they only fish during the warmest times of the day

c. they must know the right time of year for Gathering certain foods

d.) they still catch walleye with the steering method ​

Answers

The Ojibwe people pay close attention to the seasons in order to know right

time of year for gathering certain foods.

The Ojibwe mostly hunt for fishes through the use of various techniques

such as:

  • Fishing at night
  • Use of flashlight

Why do they hunt for Fishes at night?

They hunt for fishes at night because they are usually docile during that time

which enables them to catch them easily as against during the day when

they are much active.

Read more about Ojibwe people here brainly.com/question/24963033

Answer:C. They must know the right time of year for gathering certain foods

Explanation:

I got it correct