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3 years ago

What has affected almost every aspect of modern sports?

Physics

1 answer:

Semenov [28]3 years ago

6 0

Technology has affected almost every aspect of modern sports.

Answer: Option A

<u>Explanation:</u>

Modern sports consists of outdoor games and games related to having physical exercise of the body. But nowadays, technology had made the outdoor games as video games and these video games are so realistic to view that it makes the player addicted and attracted to the video games.

So in this technology, the player should use less or no physical strength to win a video games related to modern sports instead it is played on smart phones, laptops or other visible devices.

Thus, the environment of playing in a group and doing physical work for playing different games are mostly affected with rise in technology which produced realistic video games. Thus technology affects the aspects of modern sports.

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A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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3 years ago

A 55.0-g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum? Express

andrew11 [14]

Answer:

Final temperature of the aluminum = 41.8 °C

Explanation:

We have the equation for energy

E = mcΔT

Here m = 55 g = 0.055 kg

ΔT = T - 27.5

Specific heat capacity of aluminum = 921.096 J/kg.K

E = 725 J

Substituting

E = mcΔT

725 = 0.055 x 921.096 x (T - 27.5)

T - 27.5 = 14.31

T = 41.81 ° C = 41.8 °C

Final temperature of the aluminum = 41.8 °C

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3 years ago

What type of subsurface material is able to store groundwater?

notka56 [123]

<span>Porous material has many spaces that can hold(store) groundwater.</span>

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3 years ago

Suppose a constant net force of 345 N is applied to slide a heavy stationary couch across the

ira [324]

Answer:

517.5Ns

Explanation:

F=(MV - MU)/t

where MV - MU is the change in momentum,

therefore, MV - MU = Ft

= 345 X 1.

= 517.5Ns

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3 years ago

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

jeka57 [31]

Answer:

a. $3.1 m/s^2$

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

$mg sin \theta - \mu_k R = ma$

where :

m = 6.0 kg is the mass of the box

g = 9.8 m/s^2 the acceleration of gravity

$\theta=39^{\circ}$ is the angle of the slope

$\mu_k = 0.40$ is the coefficient of friction

R is the normal reaction

a is the acceleration

- Along the perpendicular direction:

$R-mg cos \theta =0$

From the 2nd equation, we get an expression for the reaction force:

$R=mg cos \theta$

And substituting into the 1st equation, we can find the acceleration:

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2$

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3 years ago