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2 years ago

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A metal bar has a volume of 32 cm3. The mass of the bar is 256 g. What is the density of the metal? A. 290 g/cm3 6 B. 8,200 g/cm

C. 8.0 g/cm3 O D. 220 g/cm

1 answer:

lina2011 [118]2 years ago

7 0

Answer:

espera que te sirva

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Estimate the average power output of the sun, given that about 1350 w/m2 reaches the upper atmosphere of the earth.

fenix001 [56]

the <u>average powe</u><u>r</u> output of the sun is <u>3.8 × </u> $10^{26}$ <u> W</u> given that about 1350 w/m2 reaches the upper atmosphere of the earth.

As we know intensity falling over a surface is equal to Energy falling per unit area per unit volume.

I = $\frac{Energy}{area*time}$

Since Power is equal to energy/time

I = power/area

or

Power = Intensity × area

Now, Distance from Sun to Earth , r = 149.6 × $10^{9}$ m

So, Surface area of the sphere of radius is:

A = 4 π $r^{2}$

= 4 × 3.14 × (149.6 × $10^{9}$ )²

= 2.81 × $10^{23}$ m²

Thus Average Power output of the sun will be:

P = 1350 × 2.81 × $10^{23}$

= 3.7935 × $10^{26}$

P = 3.8 × $10^{26}$ W

So the average power output of the sun is 3.8 × $10^{26}$ W.

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2 years ago

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A sound wave has a frequency of 741 hz in air a wavelength of 0.47m what is the temperature of the air as soon as the below city

nekit [7.7K]

Answer:

Temperature when sound wave with wavelength 0.47m has frequency 741 hz is 188 degrees Centigrade.

Step-by-Step Explanation:

Given:

Speed of sound at 0 degrees centigrade = 235 ms

frequency = f = 741 hz

wavelength = w = 0.47m

speed of sound wave = wavelength * frequency = 741 * 0.47 = 348.27 m/s

Using the formula for speed of sound at a specific temperature:

Speed of sound at T degrees Centigrade = Speed of sound at 0 degrees + 0.6 * T

When speed of sound is 348 m/s, the temperature is given by:

348 = 235 + 0.6(T)

T = (348-235)/0.6

T = 188 degrees Centigrade

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3 years ago

A boat crosses a river of width 229 m in which the current has a uniform speed of 0.756 m/s. The pilot maintains a bearing (i.e.

Stolb23 [73]

Answer:

1.85 m/s

102.438 m

Explanation:

From vector motion

The resultant velocity

$R=\sqrt{0.756^2+1.69^2}\\\Rightarrow R=1.85\ m/s$

The magnitude of speed of the speed of the boat relative to a stationary shore observer is 1.85 m/s

The distance covered is 229 m

Time = Distance / Speed

$\text{Time}=\frac{229}{1.69}\\\Rightarrow \text{Time}=135.50\ s$

Distance = Speed of current × Time

Distance = 0.756×135.5 = 102.438 m

The boat will be 102.438 m from the initial position is the boat when it reaches the opposite shore

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4 years ago

Like the moon's orbit, Earth's orbit is not a perfect circle, but an oval How do you think this might affect Earth's tides?

ExtremeBDS [4]

During the parts in the orbit where the moon is farthest away from the earth the tides will be low. Whereas during the parts where the moon is closer to the earth the tides will be higher.

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3 years ago

A merry go round exerts a force of 1000 N on a rider on the

SVEN [57.7K]

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

The given parameters are;

The force the merry go round exerts on the rider = 1000 N

The time it takes the merry go round to make one complete revolution = 15 seconds

The weight of the person = 750 N

The radius of the circle made by the person on the merry go round = r

We have;

$F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r$

Where;

m = The mass of the person

v = The velocity of the person

$F_c$ = The centrifugal force acting on the person = 1,000 N

r = The radius of the circle made by the person on the merry go round

ω = Angular velocity = 2·π/15 rad/s

We have;

The mass of the person = The weight/(The acceleration due to gravity, g)

∴ The mass of the person = 750/9.81 ≈ 76.45 kg

By substituting the calculated and known values into the equation for the centripetal force, we have;

$F_c$ = m × ω² × r

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r = 1000/(76.45 × (2·π/15)²) = 74.55 m

The radius of the circle made by the person on the merry go round = r = 74.55 m.

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3 years ago