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motikmotik
3 years ago
5

What is the direction of the magnetic field around a wire carrying a current perpendicularly into this page

Physics
1 answer:
UNO [17]3 years ago
3 0

Answer:

Clockwise direction

Explanation:

In a case of a wire carrying a current, the right hand rule is used.

The thumb in the direction of current while the finger curl around in the direction of the magnetic field.

The right hand rule applies to a current in a straight line wire.

If the direction a wire carrying a current perpendicularly into this page, the direction of the magnetic field will be in a clockwise direction .

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The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as , where a
Misha Larkins [42]

Answer:

3 times louder

Explanation:

The Loudness in decibel Db  L = 10㏒(I/I₀)  where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².

Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹

and I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₁ = 10㏒(I₁/I₀)

L₁ = 10㏒(10⁻⁹/10⁻¹²)

L₁ = 10㏒(10³)

L₁ = 3 × 10㏒10

L₁ = 30㏒10

L₁ = 30 dB

Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³

So, substituting the variables into the equation, we have

L₂ = 10㏒(I₁/I₀)

L₂ = 10㏒(10⁻³/10⁻¹²)

L₂ = 10㏒(10⁹)

L₂ = 9 × 10㏒10

L₂ =90㏒10

L₂ = 90 dB

So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3

So, Braylee's music is 3 times louder than Jessica's music

6 0
3 years ago
A 0.03 kg golf ball is hit off the tee at a speed of 34 m/s. The golf club was in contact with the ball for 0.003 s. What is the
Liula [17]

Answer:

The average force on ball by the golf club is 340 N.

Explanation:

Given that,

Mass of the golf ball, m = 0.03 kg

Initial speed of the ball, u = 0

Final speed of the ball, v = 34 m/s

Time of contact, \Delta t=0.003\ s

We need to find the average force on ball by the golf club. We know that the rate of change of momentum is equal to the net external force applied such that :

F=\dfrac{\Delta p}{\Delta t}\\\\F=\dfrac{mv-mu}{\Delta t}\\\\F=\dfrac{mv}{\Delta t}\\\\F=\dfrac{0.03\ kg\times 34\ m/s}{0.003\ s}\\\\F=340\ N

So, the average force on ball by the golf club is 340 N.

4 0
4 years ago
Answer lol! pls! !!!
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Answer:

Confused

Explanation:

more details

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3 years ago
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