One of the leading causes of permanent damage leading to hearing impairment is _____.fluid in the ears
middle ear infections
loud noise that damages hair cells
swelling in the ear canal

The amount of joules of energy used by the iron is 2.16 × J.

How to calculate energy?

Energy refers to the capacity to do work. The formula for electrical energy is as follows:

Energy = Power x Time

Where:

• Power is typically given in Watts (like a light bulb)
• Time is usually given in seconds
• Energy is usually measured in joules.

According to this question, Nicholas spends 20 minutes ironing shirts with his 1,800-watt iron. The energy used by the iron can be calculated as follows:

Energy = 1800watts × 1200s = 2.16 × J

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Answer:

Change in gravitational potential energy of the chair is,

Explanation:

It is given that,

Mass of the chair, m = 12 kg

Distance of the second floor from the ground, h = 3.3 m

The position of ground at h = 0

Let is the change in gravitational potential energy of the chair. Initial potential f the chair is equal to 0.

or

So, the change in the gravitational potential energy of the chair is 390 joules.

Final answer:

The change in gravitational potential energy of the chair is +385 J.

Explanation:

The change in gravitational potential energy of the chair can be calculated using the formula ΔPE = mgh, where m is the mass of the chair, g is the acceleration due to gravity, and h is the change in height. In this case, the mass of the chair is 12 kg, the acceleration due to gravity is 9.8 m/s², and the change in height is 3.3 m. Plugging these values into the formula, we get ΔPE = (12 kg)(9.8 m/s²)(3.3 m) = 384.72 J. Therefore, the change in gravitational potential energy of the chair is approximately +385 J.

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The direction of motion for aseries of wave fronts is represented with a ray. A ray is a line extendingoutward from the source representing the direction of propagation of the waveat any point along it. They are perpendicular to the wave fronts.

Answer:

The direction of motion for a series of wave fronts is represented with a(n) ray

:)

Answer:

a) Q1= Q2= 11.75×10^-6Coulombs

b) Q1 =15×10^-6coulombs

Q2 = 38.75×10^-6coulombs

Explanation:

a) For a series connected capacitors C1 and C2, their equivalent capacitance C is expressed as

1/Ct = 1/C1 + 1/C2

Given C1 = 3.00 μF C2 = 7.75μF

1/Ct = 1/3+1/7.73

1/Ct = 0.333+ 0.129

1/Ct = 0.462

Ct = 1/0.462

Ct = 2.35μF

V = 5.00Volts

To calculate the charge on each each capacitors, we use the formula Q = CtV where Cf is the total equivalent capacitance

Q = 2.35×10^-6× 5

Q = 11.75×10^-6Coulombs

Since same charge flows through a series connected capacitors, therefore Q1= Q2=

11.75×10^-6Coulombs

b) If the capacitors are connected in parallel, their equivalent capacitance will be C = C1+C2

C = 3.00 μF + 7.75 μF

C = 10.75 μF

For 3.00 μF capacitance, the charge on it will be Q1 = C1V

Q1 = 3×10^-6 × 5

Q1 =15×10^-6coulombs

For 7.75 μF capacitance, the charge on it will be Q2 = 7.75×10^-6×5

Q2 = 38.75×10^-6coulombs

Note that for a parallel connected capacitors, same voltage flows through them but different charge, hence the need to use the same value of the voltage for both capacitors.