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3 years ago

The car has a mass of 0·50 kg. The boy

Physics

1 answer:

PIT_PIT [208]3 years ago

7 0

Answer:

A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.

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A student is considering doing a complete repeated measures design experiment involving motor skills. The student's advisor has

Lelu [443]

Answer:

c. ABBA counterbalancing 

Explanation:

The student should not use the method because it is a progressive error management technique for each subject by introducing all treatment circumstances twice, first in one sequence, then in the other (AB, BA) by subject counterbalancing.

If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.

3 0

3 years ago

2. Compare and Contrast A fault cuts through

Elenna [48]

Explanation:

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts through. The fault labeled "E" cuts through all three sedimentary rock layers (A, B,and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen and known of.

6 0

3 years ago

In which scenario will the two objects have the greatest gravitational force

Brilliant_brown [7]

Answer: I think the answer C

Explanation:

7 0

3 years ago

Read 2 more answers

Two sound waves (speed 343 m/s) have different wavelengths. The first has a wavelength of 5.72 m, and the second a wavelength of

lys-0071 [83]

Answer:

The beat frequency is 30 Hz

Explanation:

Given;

velocity of the two sound waves, v = 343 m/s

wavelength of the first wave, λ₁ = 5.72 m

wavelength of the second wave, λ₂ = 11.44 m

The frequency of the first wave is calculated as follows;

F₁ = v/λ₁

F₁ = 343 / 5.72

F₁ = 59.97 HZ

The frequency of the second wave is calculated as follows;

F₂ = v/λ₂

F₂ = 343 / 11.44

F₂ = 29.98 Hz

The beat frequency is calculated as;

Fb = F₁ - F₂

Fb = 59.97 HZ - 29.98 Hz

Fb = 30 Hz

6 0

3 years ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

Explanation:

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

3 years ago