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ankoles [38]
3 years ago
12

Where is the frequency of infrasound in relation to the range of a human's ability to hear?

Physics
1 answer:
zhenek [66]3 years ago
3 0

Answer:

D

Explanation:

The hearing range is 20-20,000 Hz above 20,000 Hz is called ultrasonic sound and below 20 Hz is called infrasound

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When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some poi
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Answer: The force of kinetic friction is smaller than that of static friction, but F_g  remains the same.

Explanation:

The situation is same as when a book is pushed with an increasing force on a table; When the force is low, book doesn't move, until that under a given force starts moving, and then it goes on movement even if the force decreases a bit.

The physical explanation for this, that friction force adopts any value needed to avoid to move the object, till a limit value is achieved, called static friction force, equal to the normal force times the static friction coefficient.

Once in movement, the kinetic friction coefficient replaces the static one , and  in general is lower than the static one, so the force diminishes.

In the case of the box sliding down the board, the force that tries to move the object down the board, is the component of the weight parallel to the board, that can be showed that being equal to the weight times the sinus of the angle of the board with the horizontal, as follows:

F_g = m g sin θ

When θ increases, F_g does the same, so friction force always has the same magnitude than F_g (but opposite direction) so the box doesn't move, till that θ takes a value that produces a F_g equal to static friction force.

Beyond this limit angle, F_g (remaining the same for a given angle) is greater than the kinetic friction force, and the box slides.

In the limit, when θ=90º, sin θ =1⇒ F_g = mg, so the object is in free fall.

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Please help me with this question​
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Answer:

1. 12 V

2a. R₁ = 4 Ω

2b. V₁ = 4 V

3a. A = 1.5 A

3b. R₂ = 4 Ω

4. Diagram is not complete

Explanation:

1. Determination of V

Current (I) = 2 A

Resistor (R) = 6 Ω

Voltage (V) =?

V = IR

V = 2 × 6

V = 12 V

2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:

Voltage (V) = 12 V

Current (I) = 1 A

Equivalent resistance (R) =?

V = IR

12 = 1 × R

R = 12 Ω

a. Determination of R₁

Equivalent resistance (R) = 12 Ω

Resistor 2 (R₂) = 8 Ω

Resistor 1 (R₁) =?

R = R₁ + R₂ (series arrangement)

12 = R₁ + 8

Collect like terms

12 – 8 =

4 = R₁

R₁ = 4 Ω

b. Determination of V₁

Current (I) = 1 A

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) =?

V₁ = IR₁

V₁ = 1 × 4

V₁ = 4 V

3a. Determination of the current.

Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:

Resistor 1 (R₁) = 4 Ω

Voltage 1 (V₁) = 6 V

Current (I) =?

V₁ = IR₁

6 = 4 × I

Divide both side by 4

I = 6 / 4

I = 1.5 A

Thus, the ammeter (A) reading is 1.5 A

b. Determination of R₂

We'll begin by calculating the voltage cross R₂. This can be obtained as follow:

Total voltage (V) = 12 V

Voltage 1 (V₁) = 6 V

Voltage 2 (V₂) =?

V = V₁ + V₂ (series arrangement)

12 = 6 + V₂

Collect like terms

12 – 6 = V₂

6 = V₂

V₂ = 6 V

Finally, we shall determine R₂. This can be obtained as follow:

Voltage 2 (V₂) = 6 V

Current (I) = 1.5 A

Resistor 2 (R₂) =?

V₂ = IR₂

6 = 1.5 × R₂

Divide both side by 1.5

R₂ = 6 / 1.5

R₂ = 4 Ω

4. The diagram is not complete

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