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Lana71 [14]
3 years ago
8

What might happen if a person’s ear canal was blocked?

Physics
1 answer:
mylen [45]3 years ago
7 0

Answer:

D. Sounds would be harder to hear.

Explanation:

if the ear canal is blocked then the sound waves cannot reach the ear drum thus we cannot hear clearly.

hope this helps please mark as brainliest:)

You might be interested in
The image shows the electric field lines around two charged particles.
DENIUS [597]
It's either 3 or 4 I know this becuase I have read a book about electricity
8 0
3 years ago
Read 2 more answers
When a positive charge moves in the direction of the electric field, what happens to the electrical potential energy associated
VladimirAG [237]

Answer:

B it decreases

Explanation:

the movement of a positive test charge in the direction of an electric field would be like a mass falling downward within Earth's gravitational field. Both movements would be like going with nature and would occur without the need of work by an external force. This motion would result in the loss of potential energy

6 0
3 years ago
Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a
butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0
2 years ago
What is the best way to support your hypothesis
Diano4ka-milaya [45]

Answer:

Making a Hypothesis

Explanation:

-Research the subject of your question. Review the literature and find out as much as you can about previous information and discoveries surrounding your question.

-Develop an educated guess that answers your initial question. This is your hypothesis. Make a prediction based on your hypothesis and state it as a cause-effect relationship.

4 0
3 years ago
Q1:A large tank is filled with water. The pressure on the base of the fish tank is 4000N/m². The base of the tank is a rectangle
Ymorist [56]

Answer:

F = 36 kN

Explanation:

It is given that,

The pressure on the base of the fish tank is 4000N/m².

The base of the tank is a rectangle measuring 2.0m by 4.5m.

Area of the base of the tank is 9 m²

We need to find the force on the base caused by the base of the water. Pressure on the base of the tank is given by the force acting per unit area such that,

P=\dfrac{F}{A}\\\\F=P\times A\\\\F=4000\times 9\\\\F=36000\ N\\\\F=36\ kN

So, the force of 36 kN is acting on the base by the base of water.

7 0
3 years ago
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