What kind of electromagnetic waves are used by an FM radio?.

By what factor does the sound intensity increase if the sound intensity level increases from 60 db to 61 db?

tiny-mole [99]

Answer:

$\frac{I_{2}}{I_{1}} = 10^{0.1}$

Explanation:

The sound intensity level β can be mathematically defined as

$\beta(dB) = 10log_{10}(\frac{I}{I_{o} } )$

where $I_{o}$ is the reference intensity and is equal to $10^{-12}$ $W/m^{2}$ .

Now,

$\beta_{2} - \beta_{1} = 10 log\frac{I_{2} }{I_{o}} - 10 log\frac{I_{1} }{I_{o}}\\62-60= 10log(\frac{\frac{I_{2} }{I_{o} } }{\frac{I_{1} }{I_{o} } } )\\1=10log(\frac{I_{2} }{I_{1} } )\\\\\frac{I_{2} }{I_{1} }=10^{0.1}$

6 0

4 years ago

Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc

lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

where Q= calories transferred between the system and its environment or another system (unity: calorie or cal) (what we are trying to find out),

m= mass of the substance (unity: grams or g), and

ΔT= difference of temperature (unity: Celsius degrees or °C); and

m= 95g and ΔT= 16°C:

Q= 1 cal/(g.°C).95g.16°C =<u> 1,520.00 cal </u>

8 0

4 years ago

baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the

Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0

3 years ago

Give one example of a thermodynamic cycle that does not account for the carnot efficiency.

Arturiano [62]

Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.

The Carnot cycle is a hypothetical cycle that takes no account of entropy generation. It is assumed that the heat source and heat sink have perfect heat transfer. The working fluid also remains in the same phase, as opposed to the Rankine cycle, in which the fluid changes phase. A practical thermodynamic cycle, such as the Rankine cycle, would achieve at most 50% of the Carnot cycle efficiency under similar heat source and heat sink temperatures.

<h3>What is Thermo-Electrochemical converter?</h3>

In a two-cell structure, a thermo-electrochemical converter converts potential energy difference during hydrogen oxidation and reduction to heat energy.

It employs the Ericsson cycle, which is less efficient than the Carnot cycle. In a closed system, it converts heat to electrical energy. There are no external input or output devices.

This means there will be no mechanical work to be done, as well as no exhaust. As a result, Carnot efficiency is not taken into account in this cycle. Carnot efficiency is accounted for by other options such as turbine and engine.

Learn more about Thermo-Electrochemical converter here:

brainly.com/question/13040188

#SPJ4

4 0

2 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago