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3 years ago

8

If 2,575J of heat is added to a 250g piece of silver, the change in temperature of the silver will be 43.6 °C? What is the speci

fic heat of of silver

2 answers:

qaws [65]3 years ago

7 0

Answer:

0.236 C

Explanation:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

58.56°C

Explanation:

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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

The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and

Salsk061 [2.6K]

The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

Let:

b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°

We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.

5 0

3 years ago

How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the $(velocity)^2$ and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:

$\text { Kinetic Energy }=\frac{1}{2} m v^{2}$

Better understood from numerical example as given:

If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?

This can be solved as follows:

$\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}$

$\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}$

It shows that man A will have more K.E.

Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.

4 0

3 years ago

Light from distant galaxies most likely shows a ...red shift, indicating that the universe is expandingblue shift, indicating th

ELEN [110]

Answer:

red shift, indicating that the universe is expanding

Explanation:

Doppler effect occurs when a source of a wave (e.g. light, or sound waves) moves relative to an observer; as a result of this relative motion, the wavelength of the wave appears lengthened/shortened to the observer. Two situations can occur:

- The source of the wave is moving towards the observer - in this case, the wavelength of the wave becomes shorter. If the wave is visible light, such as the light emitted by distant galaxies, this means that the wavelength of the light shifts towards the blue-end of the spectrum (blue-shift)

- The source of the wave is moving away from the observer - in this case, the wavelength of the wave becomes longer. If the wave is visible light, such as the light emitted by distant galaxies, this means that the wavelength of the light shifts towards the red-end of the spectrum (red-shift)

In our universe, we observe a red-shift for all the distant galaxies: this means that these galaxies are moving away from us, so this is an indication that the universe is expanding.

5 0

4 years ago

Korolek [52]

Answer:

143.352 watt.

Explanation:

So, in the question above we are given the following parameters or data or information that is going to assist us in answering the question above efficiently. The parameters are:

"A 1.8 m wide by 1.0 m tall by 0.65m deep home freezer is insulated with 5.0cm thick Styrofoam insulation"

The inside temperature of the freezer = -20°C.

Thickness = 5.0cm = 5.0 × 10^-2 m.

Step one: Calculate the surface area of the freezer. That can be done by using the formula below:

Area = 2[ ( Length × breadth) + (breadth × height) + (length × height) ].

Area = 2[ (1.8 × 0.65) + (0.65 × 1.0) + (1.8 × 1.0)].

Area = 7.24 m^2.

Step two: Calculate the rate of heat transfer by using the formula below;

Rate of heat transfer =[ thermal conductivity × Area (T1 - T2) ]/ thickness.

Rate of heat transfer = 0.022 × 7.24(25+20)/5.0 × 10^-2 = 143.352 watt.

8 0

3 years ago