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4 years ago

Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss

Physics

1 answer:

vekshin14 years ago

4 0

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of transfer by

Explanation:

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What process takes solid CO2 from a solid to a gas? A) boiling B) melting C) sublimation D) vaporization

frutty [35]

This process is D) vaporization. Hope this helps! :)

<h3>CloutAnswers</h3>

7 0

4 years ago

Read 2 more answers

It is 4.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

Musya8 [376]

Answer:

Explanation:

To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.

But first we have to calculate the time

$t_{1}=\frac{x}{v_{1}}=\frac{4km}{10\frac{km}{h}}=0.4h=0.4(3600s)=1440s\\t_{2}=\frac{x}{v_{2}}=\frac{4km}{3\frac{km}{h}}=1.3h=1.3(3600s)=4800s\\$

Now we use E=W*t for both times

$E_{1}=t_{1}W_{1}=(1440s)(700W)=1008000J\\E_{2}=t_{2}W_{2}=(4800s)(290W)=1392000J\\$

A. Hence, by running the energy consumed is lower.

E1=1008000J

E2=1392000J

C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.

5 0

3 years ago

What are the coordinates of the point that is 1/5 of the way from A(-7,-4) to B(3, 6)?

Olegator [25]

Answer:

coordinate will be

r = (1.33,4.33)

Explanation:

As we know that if a point will divide two given coordinates in m : n ratio

then in that case the point is given as

$x = \frac{mx_1 + nx_2}{m+n}$

$y = \frac{my_1 + ny_2}{m+n}$

here we know that two points are

(-7, -4) and (3, 6)

and the ratio is given as 1 : 5

now from above formula we will have

$x = \frac{1(-7) + 5(3)}{1 + 5} = 1.33$

$y = \frac{1(-4) + 5(6)}{1 + 5} = 4.33$

so coordinate will be

$r = (1.33, 4.33)$

7 0

4 years ago

The waste products of a nuclear fusion power plants can be best described as

I am Lyosha [343]

They can be described as small in quantity and very dangerously radioactive.

4 0

3 years ago

Read 2 more answers

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

$I_1 = \frac{I_p}{2}$

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

3 years ago