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Lorico [155]
3 years ago
14

The seven sciences the study of kinesiology includes are: exercise physiology, exercise anatomy, biomechanics, exercise psycholo

gy, exercise sociology, motor learning, and sport pedagogy. Question 3 options:
True False
Physics
1 answer:
Alex3 years ago
5 0

Answer:

True

Explanation:

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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
4 years ago
20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu
Anna71 [15]

Answer:

B. V_{f}= 10\,cubic\,inches

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

PV=nRT

With T the temperature, V the volume, P the pressure, R the perfect gas constant and n the number of mol, we are going to use the subscripts i for the initial state when the gas has 20 cubic inches of volume and absolute pressure of 5 psi, and final state when the gas reaches 10 psi, so we have two equations:

P_{i}V_{i}=n_{i}RT_{i} (1)

P_{f}V_{f}=n_{f}RT_{f} (2)

Assuming the temperature and the number of moles remain constant (number of moles remain constant if we don't have a leak of gas) we should equate equations (1) and (2) because T_{i}=T_{f}, n_{i}=n_{f} and R is an universal constant:

P_{i}V_{i}= P_{f}V_{f}, solving for V_{f}

V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}

V_{f}= 10 cubic\,inches

6 0
3 years ago
How can you increase the potential energy of a bouncing ball
ad-work [718]

Answer:

When you lift the ball, you are doing work to increase its gravitational potential energy. When you then release the ball, gravitational energy is transformed into kinetic energy as the ball falls. When the ball hits the floor, the ball's shape changes as it flattens against the floor.

Explanation:thats should be the way^^ in explaining

7 0
3 years ago
A stone is projected from the ground with a velocity 14m/s one second later it clears a wall 2 m high. The angle of projection i
Eva8 [605]

Answer:

30 degrees

i can't put in my work on paper

6 0
3 years ago
Read 2 more answers
If the electron is initially in the ground level, E1=0.625E[infinity]E1=0.625E[infinity], and absorbs a photon, what maximum wav
tekilochka [14]

Answer:

 λ = 3.1824 10-25 m

Explanation:

To release the electron from the power well the absorbed photon electro must be greater than or equal to the energy of the electron in the power well

           

The photon energy is E_photon = 0.625 J

The speed of light is

       c = λ f

         

Let's use Planck's equation

        E = h f

             

We substitute

       E = h c / λ

        λ = h c / E

        λ = 6.63 10-34 3 108 / 0.625

         λ = 3.1824 10-25 m

4 0
3 years ago
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