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3 years ago

14

AYUDAAAAAAA Un ciclista de montaña se mueve cuesta abajo por la pendiente, si acelera a 3 m/s2 y se demora en bajar 15 minutos (

900 segundos) ¿Qué distancia recorre? AYUDENME PORFAVOOOR

1 answer:

dezoksy [38]3 years ago

3 0

The question is: HELPAAAAAAA A mountain biker moves downhill if he accelerates at 3 m / s2 and takes 15 minutes (900 seconds) down, how far does he travel? HELP ME PLEASE

Answer: The distance travelled by the mountain biker is 2430 km.

Explanation:

Given: Acceleration = $3 m/s^{2}$

Time = 15 minutes (900 sec)

Displacement = ?

Formula used to calculate the final velocity is as follows.

$v = u + at$

where,

v = final velocity

u = initial velocity = 0 m/s (here)

t = time

a = acceleration

Substitute the values into above formula as follows.

$v = u + a \times t\\v = 0 + 3 m/s^{2} \times 900 s\\= 2700 m/s$

The relation between velocity (speed) and distance is as follows.

$Speed = \frac{Distance}{time}\\2700 m/s = \frac{Distance}{900 s}\\Distance = 2430000 m (1 m = 0.001)\\= 2430 km$

Thus, we can conclude that distance travelled by the mountain biker is 2430 km.

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a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

Now we can find the radiation intensity with the equation

$I=\frac{P}{A}$

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

Learn more about electromagnetic radiation:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

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