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3 years ago

When measurements are close to their true values, they are said to be

Physics

2 answers:

Nikitich [7]3 years ago

8 0

Answer:

A) accurate

Explanation:

Accuracy refers to how close a measurement is to the true value of what is being measured.

katovenus [111]3 years ago

3 0

Answer:

When measurements are close to true

value they are called to be accurate

Explanation:

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What happens to the ball's velocity while the ball is traveling upwards?

Bess [88]

If the ball does not have a propeller or jet engine on it, then it is an object
in free fall. That means its downward speed grows by 9.8 m/s for every
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If it happens to be traveling upward at the moment, then that won't last long.
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7 0

3 years ago

A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experim

xxTIMURxx [149]

Answer:

Placing cells in series increases the

voltage in the circuit by 1.5 V for each

cell. Increasing the voltage increases the

brightness of the bulb

Explanation:

I. MAKING THE CONNECTION

How many terminals are located on the battery? 2

How many terminals are located on the bulb? 2

II. PLACING CELLS IN SERIES

What is the effect on the brightness of the bulb by increasing the number of cells?

The bulbs become brighter when increasing the number of cells.

What changes occur in the current in the circuit when increasing the number of cells?

Increasing the number of cells increases the current in the circuit.

What changes occur in the voltage in the circuit when increasing the number of cells?

Increasing the number of cells increases the voltage (for cells in series the voltage is additive).

What changes occur in the resistance in the circuit when increasing the number of cells?

The resistance is determined by the number of bulbs. The resistance in the circuit remains unchanged.

III. PLACING BULBS IN SERIES

What is the effect on the brightness of the bulbs by increasing the number of bulbs?

Increasing the number of bulbs decreases the brightness of the bulbs.

What changes occur in the resistance in the circuit as more bulbs are added?

The resistance increases. In a series circuit, adding bulbs increases the resistance in the circuit.

What changes occur in the current in the circuit as more bulbs are added?

Increasing the resistance decreases the current.

Observations on unscrewing one bulb. Explain your observations.

A complete circuit requires the electrons to move from the negative terminal of the battery to the

positive terminal. When one bulb is unscrewed it opens the circuit preventing a complete circuit and

the electrons cannot return to the battery.

IV. PLACING BULBS IN PARALLEL

Compare the brightness with two bulbs in parallel with two bulbs in series.

Two bulbs in parallel are brighter than two bulbs in series.

How does increasing the number of circuits (bulbs) change the current and resistance?

In a parallel circuit each bulb is in its own circuit. As bulbs are added the resistance in the circuit

decreases since each circuit is another pathway for electrons to move from one end of the circuit

4 0

4 years ago

A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m

Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0 (1)

2.070k + Φ = 70.53. (2)

Subtracting (2) -(1)