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ahrayia [7]
3 years ago
8

The motion of a free falling body is an example of __________ motion​

Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

uniformly accelerated motion

Explanation:

The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.

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Alloys are mixtures of ________.
Hunter-Best [27]

Answer:

e.metals

Explanation:

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4 years ago
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A 0.20 mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to
olga55 [171]

Answer:

d

Explanation:

Ya gon find the Kenitic Energy first

K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004

and now the replacement:

0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

4 0
3 years ago
Is there any induced current in the pickup coil when the polarity of the bar magnet is flipped?
stiks02 [169]

Answer:

Faraday's law , he direction of the magnetic field changes by 180º, in the polarity inversion processes, induces a voltage.  

Explanation:

For this exercise let's use Faraday's law

          E = - dФ / dt

          Ф = B.A = B A cos θ

where B is the magnetic field, A is the area and θ is the angle between the field line and the normal to the area.

We can see that an electromotive force (voltage) is indexed when there is a variation of the field B, a variation of the area and change of the angle or when there is a combinational of them.

In this case, the magnitude of the field is constant, as the wire is rigid metal, the area is constant, but the direction of the magnetic field changes by 180º, in the polarity inversion processes, for which reason each change induces a voltage.

If a voltage is created in the ring, which has a resistance, a current is also generated in it.

Therefore the answer is If a current is created in the hoop

8 0
3 years ago
A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
rjkz [21]

Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

4 0
3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

Δu=1300kJ/kg  

Explanation:

Energy at the initial state

p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)

Is saturated vapor at initial pressure we have

p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)

Process 2-3 is a constant volume process

p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg

Δu=1300kJ/kg  

3 0
3 years ago
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