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kirill115 [55]
2 years ago
14

Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if R

oss takes two months off from training what will likely occur?
Physics
1 answer:
damaskus [11]2 years ago
4 0

If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.

  • In as little as 3–4 weeks after beginning strength training, Ross will probably experience weight increase, energy loss, diminished balance, diminished strength (making it tougher to carry out daily tasks), and overall fewer fitness levels.
  • Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.
  • A decrease in strength and muscle mass, with beginners experiencing a smaller decline in strength than experienced lifters.
  • Ross will experience Increased VO2 Max from exercise. VO2 Max is almost completely lost in people who train at lower intensities.

learn more about fitness here: brainly.com/question/13490156

#SPJ10

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Allen Aby sets up an Atwood Machine and wants to find the acceleration and the tension in the string. Please help him. Two block
Vitek1552 [10]

<u>Answers:</u>

In order to solve this problem we will use Newton’s second Law, which is mathematically expressed after some simplifications as:

<h2>F=ma   (1) </h2>

This can be read as: The Net Force F of an object is equal to its mass m multiplied by its acceleration a.

We will also need to <u>draw the Free Body Diagram of each block</u> in order to know the direction of the acceleration in this system and find the Tension T of the string (<u>See figure attached).  </u>

We already know<u> m_{2} is greater than m_{1}</u>, this means the weight of the block 2 P_{2} is greater than the weight of the block 1 P_{1}; therefore <u>the acceleration of the system will be in the direction of P_{2}</u>, as shown in the figure attached.

We also know by the information given in the problem that <u>the pulley does not have friction and has negligible mass</u>, and <u>the string is massless</u>.

This means that the tension will be the same along the string regardless of the difference of mass of the blocks.

Now that we have the conditions clear, let’s begin with the calculations:

1) Firstly, we have to find the weight of each block, in order to verify that block 2 is heavier than block 1.

This is done using equation (1), where the force of the weight P is calculated using the <u>acceleration of gravity</u> g=9.8\frac{m}{s^{2}}  acting on the blocks:


<h2>P=mg   (2) </h2>

<u>For block 1: </u>

P_{1}=m_{1}g   (3)

P_{1}=1.5kg(9.8\frac{m}{s^{2}})    

<h2>P_{1}=14.7N   (4) </h2>

<u>For block 2: </u>

P_{2}=m_{2}g   (5)

P_{2}=2.4kg(9.8\frac{m}{s^{2}})    

<h2>P_{2}=23.52N      (6) </h2>

Then, we are going to <u>find the acceleration a of the whole system: </u>

F_{r}=P_{1}+P_{2}   (7)

<h2>P_{1}+P_{2}=(m_{1}+m_{2})a   (8) </h2>

Where the Resulting Force F_{r}  is equal to the sum of the weights P_{1} and P_{2}.  

In the figure attached, note that P_{1} is in opposite direction to the acceleration a, this means it must <u>have a negative sing</u>; while P_{2} is in the same direction of a.

Here we only have to isolate a from equation (8) and substitute the values according to the conditions of the system:

-14.7N+23.52N=(1.5kg+2.4kg)a  

8.82N=(3.9kg)a  

Then:

a=\frac{8.82N }{3.9kg}  

<h2>a=2.26\frac{m}{ s^{2}}  </h2><h2>This is the acceleration of the system. </h2>

2) For the second part of the problem, we have to find the tension T of the string.

We can choose either the Free Body Diagram of block A or block B to make the calculations, <u>the result will be the same</u>.  

Let’s prove it:

For m_{1}

we see in the free body diagram that the <u>acceleration is in the same direction of the tension of the string</u>, so:

F_{r}=T-P_{1}   (9)

T-P_{1}=m_{1}a   (10)

T-14.7N=(1.5kg)( 2.26\frac{m}{ s^{2}})    

Then;

<h2>T=18.09N   This is the tension of the string </h2><h2> </h2>

For m_{2}

we see in the free body diagram that the acceleration is in opposite direction of the tension of the string and must <u>have a negative sign,</u> so:

F_{r}=T-P_{2}   (9)

T-P_{2}=m_{2}a   (10)

T-23.52N=(2.4kg)(-2.26\frac{m}{ s^{2}})    

Then;

<h2>T=18.09N    This is the same tension of the string </h2>

6 0
3 years ago
Match the term with its description.
Mrac [35]

A=Ion

B=pH

C=Acid

D=Alkaline

6 0
3 years ago
12.
N76 [4]
I think it is D. Brownies are harder to bake than cupcakes. an example of an objective statement is why a company is hiring for that certain job. let me know if this is correct!
Hope this helped
5 0
3 years ago
Read 2 more answers
The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a
SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm=1900\times 10^{-9}m

Where 1nm=10^{-9} m

\lambda_2=400nm=400\times10^{-9}m

\lambda_1=700nm=700\times 10^{-9}m

We have to find  the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

\theta=sin^{-1}(\frac{m\lambda}{d})

Using the formula

m=1

\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})

\theta=21.62^{\circ}

Now, m=2

\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})

\theta_2=24.90^{\circ}

\Delta \theta=\theta_2-\theta_1

\Delta \theta=24.90-21.62

\Delta \theta=3.28^{\circ}

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0
3 years ago
A load is to be pulled up an inclined plane using a pulley system. The inclined plane is 25.0 ft long and 5.00 ft high. The pull
tangare [24]
Advantage equals 0.1, mate
1) advantage of plane: 5 divided by 25 = 0.2
2) advantage of fixed pulleys = 0
3) advantage of movable pulley = 0.5
4) and you multiply 0.2 and 0.5 = 0.1

to lift 1kg you don't need force 10N, you will just need 1N
4 0
4 years ago
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