Which of the following is a benefit of physical activity?

Complete Question  

The complete question is shown on the first and second uploaded image  

Answer:

When water is used the reading is  [tex]    R =  2281.6 \  cm  [/tex]

When mercury is used the reading is [tex]   R =  23.83 \ cm [/tex]  

The best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Explanation:

From the question we are told that  

    The length of the leg of the manometer to the top of the tank is  d =  500cm

   The toluene level where in the tank where the height of the manometer fluid level in the open arm is equal to the height where the manometer is connected to the tank is  h =150 cm  

   The manometer reading is  R  

Generally at the point where the height of the open arm is equal to the height of the of the point connected to the tank ,  

   The pressure at the height of the both arms of the manometer corresponding to the base of the tank are equal  

     i.e   [tex]P_1 = P_2[/tex]

Here [tex]P_1[/tex] is the pressure of the manometer at the point corresponding to the base of the tank and this is mathematically represented as  

       [tex] P_{atm} + P_1 =  P_{atm} + P_t[/tex]

Here [tex]P_t[/tex] is the pressure due to the toluene level in the tank and in the arm of the manometer connected to the tank and this is mathematically represented as  

         [tex]P_t  =  \rho_t  * g  * h_i[/tex]

Here  

      [tex]\rho_t [/tex] is the density of toluene with value  [tex]\rho_t =  867 kg/m^3 [/tex]

 [tex]h_i[/tex] is the height of the connected arm above the point equivalent to the base of the tank , this mathematically represented as

          [tex]h_i =  d - h + R[/tex]

  and  [tex] P_2 [/tex] is the the pressure at the open arm of the manometer at the point equivalent to the base of the base of the tank and this is mathematically represented as

         [tex] P_2 =  \rho_f * g *  h_f [/tex]

Here  

       [tex]\rho_f[/tex] is the density of the fluid in use , if it is water the density is  

       [tex]\rho_w =  1000 \  kg /m^3 [/tex]

and  if it is  mercury the density is  

        [tex]\rho_m =  13600 \  kg /m^3 [/tex]

[tex]h_f[/tex] is the height of the  fluid in the open arm of the manometer from the point equivalent to the base of the tank which is equivalent the manometer reading R

So when the fluid is water we have

      [tex] P_{atm} +  \rho_t* g *(d - h + R) =  P_{atm} + \rho_f * g *  h_f[/tex]

=>   [tex]   \rho_t* (d - h + R) =   \rho_w *  h_f[/tex]        

=>   [tex]    867  (500 - 150 + R) =    1000 *  R [/tex]

=>    [tex]    R =  2281.6 \  cm  [/tex]

So when the fluid is mercury we have      

  [tex]   \rho_t* (d - h + R) =   \rho_m *  h_f[/tex]          

=>   [tex]    867  (500 - 150 + R) =   13600  *  R [/tex]  

=>   [tex]   R =  23.83 \ cm [/tex]  

The difference in the mercury reading for mercury due to the fact that they have different densities as we have seen in this calculation

So the best fluid to use is mercury because for water a slight change in toluene level will cause a  large change in height .

Answer:

5N westward, acting on the student

Explanation:

Answer: 5N westward, acting on the student

Explanation:

Answer:

2991.47 [cm^2]

Explanation:

To solve this problem we must perform a dimensional analysis and use the corresponding conversion values:

[tex]3.22[ft^{2}]*\frac{12^{2}in^{2} }{1^{2}ft^{2}} *\frac{2.54^{2}cm^{2} }{1^{2}in^{2} } \\2991.47[cm^{2}][/tex]

Answer:

A. 50 km south

Explanation:

hope this helps!

Answer:

8.8 hours

Explanation:

plz give me a Brainliest

Answer:

This has 5 significant figures.

Answer:

30 J

Explanation:

height = 3 m

mass = 1kg

acceleration due to gravity = 10m/s²

Potential Energy = mass× gravity × height

= 1kg ×10m/s²× 3m

= 30 J

Answer:

D

Explanation:

I could be wrong, But I'm pretty sure it's D

Answer:

UR ANSWER IS WAVELENGTH

===========================================================

Explanation:

Consider a mass of 10 kg, so m = 10

Let's say we apply a net force of 20 newtons, so F = 20

The acceleration 'a' is...

F = ma

20 = 10a

20/10 = a

2 = a

a = 2

The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.

---------------

Now let's double the net force on the object

F = 20 goes to F = 40

m = 10 stays the same

F = ma

40 = 10a

10a = 40

a = 40/10

a = 4

The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.

---------------

In summary, if you double the net force applied to the object, then the acceleration doubles as well.

Acceleration is directly proportional to the net force on an object, and inversely proportional to its mass.

So if an object's mass stays the same while the net force on it doubles, then its acceleration will also double.

We don't know anything about the "trials".  This sounds like it might be a follow-up to a lab experiment that was performed when we weren't there.

We also don't know anything about "question 1".

Answer:

Δx = -80m

Explanation:

The object's displacement is -80m because it landed 80m below where it started.

Answer: Approximately 302 m/s^2

================================================

Work Shown:

s = starting velocity = 0

f = final velocity  = 46

d = distance = 3.5

a = acceleration = unknown (we're solving for this)

[tex]f^2 = s^2 + 2a*d \ \ \text{ ..... one of the kinematics equations}\\\\46^2 = 0^2 + 2a*3.5\\\\2116 = 7a\\\\7a = 2116\\\\a = \frac{2116}{7}\\\\a \approx 302.28571\\\\a \approx 302[/tex]

The acceleration to three sig figs is roughly 302 m/s^2

The acceleration is so large because the ball's final velocity is incredibly fast in such a short amount of time.

Answer:

Explanation:

Time taken by jogger to travel the distance to finishing line = 2.1 / 7.3

= .28767 hr

Bird will keep flying for this time period

distance covered by bird = speed x time

= 29.2 x .28767 km

= 8.4 km .

Explanation:

Acceleration is the rate of change in the speed of an object. If initial and final speed are v₁ and v₂. It changes speed in time t. Its formula is given by :

[tex]a=\dfrac{v_2-v_1}{t}[/tex]

The problem is "A skater increases her velocity from 2.0 m/s to 10.0 m/s in 3.0 seconds. What is the skater’s acceleration? "

In this problem,

v₁ = 2 , v₂ = 10 m/s and t = 3 s

Acceleration of the object,

[tex]a=\dfrac{10-2}{3}\\\\a=2.67\ m/s^2[/tex]

So, the acceleration of the skater is [tex]2.67\ m/s^2[/tex].

Answer:

1305 I think but not sure

Answer:

It's 31

Explanation:

add 15 and 16

Answer:

True

Explanation:

The term 'plyometrics' is used interchangeably with the term 'jump training'. The technique can be used for training in sports that require explosive movements.

Answer:

162

Explanation:

When the amplitude of the wave is 8 units, the corresponding energy of the wave is 128 units. Therefore, option C is correct.

The term amplitude is defined as the maximum distance moved by a point on a vibrating body or wave calculated from its equilibrium position.

We have given,

first amplitude, A = 1, corresponding energy = 2

second amplitude, A = 2, corresponding energy = 8

The energy of a string at  a given amplitude is calculated as follows;

E ∝ A²

Energy varies as the square of the change in amplitude.

If amplitude of one has energy of two

Amplitude of 2 (double of 1)

= ( 2 x 2 ) = 4 x 2

= 8  4 times the energy at amplitude of 1.

Then for 8 units of amplitude change

8 x 8 = 64 x 2

= 128  

Thus, When the amplitude of the wave is 8 units, the corresponding energy of the wave is 128 units, option C is correct.

To learn more about an amplitude, follow the link;

https://brainly.com/question/15930409

#SPJ2

Your question is incomplete, most probably your question was

The data table shows how the amplitude of a mechanical wave varies with the energy it carries. Analyze the data to identify the mathematical relationship between amplitude and energy. Use your equation to find the energy if the amplitude is 8 units.

Amplitude

1

2

3

4

energy

2

8

18

32

A.66 units

B. 108 units

C. 128 units

D. 88 units ​

Answer:

Capacidad de almacenamiento ofrecida por los DVD

Hay que mencionar que es posible conseguir DVD de una o dos capas. Los DVD que cuentan con una capa son capaces de almacenar aproximadamente 4.7GB de datos, mientras que los de doble capa ofrecen la posibilidad de almacenar alrededor de 8.5GB

Explanation:

Answer:

4 layers

Explanation:

i took the quiz

Answer: The process is called artificial selection because people (instead of nature) select which organisms get to reproduce.

Explanation: This is evolution through artificial selection and the characteristics to reproduce, causing the evolution of farm stock.

Answer:

Mechanical energy

Explanation:

Mechanical energy is needed for movement of objects. Muscles convert chemical energy provided by the rest of the body to allow movement.

Answer:

a strong nuclear force is a powerful force of attraction that acts only on the neutrons and protons in nucleus

so the answer is: STRONG NUCLEAR

Answer:

False

Explanation:

More than one management style with all people is the most effective leadership technique for any organization. So, the given statement is False.

Leadership styles are defined as those adopted by company leaders to increase efficiency, satisfaction, growth and advancement in all positions in the organization. It requires a different style in different situations and each leader needs to know when to show a special behavior.

These strategies define every leader's personal leadership style. Some effective leadership Techniques are as follows:

Thus, more than one management style with all people is the most effective leadership technique for any organization. So, the given statement is False.

Learn more about Marketing techniques, here:

https://brainly.com/question/29843697

#SPJ2

Answer:

Usually, the relationship between mass and weight on Earth is highly proportional; objects that are a hundred times more massive than a one-liter bottle of soda almost always weigh a hundred times more—approximately 1,000 newtons, which is the weight one would expect on Earth from an object with a mass slightly greater ...

Answer:

the corrdct answer is core

Answer:

bada bop boop pow

Explanation:

It's gonna take a huge amount of time and effort, and we're gonna spend a huge amount of money, to build a vacuum chamber big enough to do this experiment.  

So when we're finally ready to try it, the cannonball and golf ball will make a  kind of cute experiment, but let's make it really interesting.  Let's ALSO try it with a feather, an apple, a watermelon, a tennis ball, a basketball, a coffeepot, a chair, an old computer, a set of dishes, a trombone, a bicycle, a math book, a TV set, a gallon of milk, a skateboard, a shotgun, a concrete block, a car, a waterbed, a schoolbus, a battleship, your dog, and my wife !

Drop every single one of them from a height of 10 meters.

Every single one of them accelerates at 9.8 m/s² as it falls, takes 1.429 seconds to reach the floor, and hits the floor at a speed of 14 m/s.  The feather, the schoolbus, and my wife stay together all the way down.

The only way this doesn't happen is if ...

-- you do this experiment in some other place that's not Earth, or

-- you throw one of the objects and you don't just drop it, or

-- the vacuum chamber has a leak and some air gets into it.

Because that's how gravity works.

Answer:

The correct option is;

C. The temperature of the air affects the speed of sound

Explanation:

The information given are;

The speed with which sound travels in the desert air = 358 m/s

The speed with which sound travels in the polar air = 330 m/s

The major difference between the desert air and the polar air is that the temperature of the desert air is hotter than the temperature of the air in the polar region. Therefore the speed of sound is affected by the air temperature.

The equation that gives the relationship of the speed of sound in air, [tex]v_{air}[/tex], to temperature is presented as follows;

[tex]v_{air} = 331\frac{m}{s} \times \sqrt{\dfrac{T_K}{273 K} } = 331\frac{m}{s} \times \sqrt{1 + \dfrac{T_{^{\circ}C}}{273 ^{\circ} C} }[/tex]

Which shows that the speed of sound in air, [tex]v_{air}[/tex], rises as the temperature of the air rises.

Where;

[tex]T_{^{\circ}C}[/tex] = The temperature of the air in degrees Celsius

[tex]T_K[/tex] = The temperature of the air in degrees Kelvin.