Ask question

Ask question

All categories

3 years ago

14

In which one(s) of the following situations will there be an INCREASE in Kinetic Energy? Group of answer choices A block slides

down a frictionless incline A box is pulled across a rough floor at constant speed. A stone at the end of a string is whirled in a horizontal circle at constant speed. A projectile approaches its maximum height. A merry go round rotates faster due to the push by a person.

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

The Kinetic Energy INCREASE in the following situations:

a)A block slides down a frictionless incline:

e)A merry go round rotates faster due to the push by a person.

Explanation:

The energy kinetics is proportional to the square of the velocity:

$E_k=1/2*mv^2$

We study the cases:

a)A block slides down a frictionless incline:

The gravity made a positive work and the box get a extra velocity, then the kinetics energy INCREASE

b)A box is pulled across a rough floor at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

c)A stone at the end of a string is whirled in a horizontal circle at constant speed:

The magnitude of the velocity does not change, so the kinetics energy does not change as well

d)A projectile approaches its maximum height:

If the projectile approaches its maximum height, its velocity approaches to zero. Then the kinetics Energy DECREASE

e)A merry go round rotates faster due to the push by a person.

Thanks to the push, the magnitude of the velocity INCREASE, so the kinetics energy INCREASE as well

You might be interested in

A 10kg block is attached to a light cord that is wrapped around the pulley of an electric motor. What is the power output of the

Scorpion4ik [409]

Answer:

Option B is correct.

Power = 360 W

Explanation:

Power = Work done/time

Work done = Force × distance moved through by the force

Power = Force × (distance moved through by the force/time)

(Distance moved through by the force/time) = velocity = 3 m/s

Power = Force × velocity

Force = ma

But the acceleration in this case is this acceleration + acceleration due to gravity because the force has to be overcoming the force of gravity to now move the object upward at 2 m/s²

a = (2 + g) (assume acceleration due to gravity = 10 m/s²

a = 2 + 10 = 12 m/s²

F = ma = 10 × 12 = 120 N

Power = F × v = 120 × 3 = 360 W

4 0

3 years ago

The four qualifications for the presidency are outlined in the __________. A. US Constitution B. Presidential Charter C. Declara

mojhsa [17]

The four qualifications for the presidency are outlined in the constitution.

5 0

3 years ago

Read 2 more answers

elixir [45]

Answer:Squinting is often a sign of vision

problems. The result of squinting the

eye will result in a force being aplied

that

Explanation:

7 0

3 years ago

A car moving south speeds up from 10m/s to 40m/s in 15 seconds what is the cars acceleration

torisob [31]

The acceleration of the car will be 2m/s².Option A is correct. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

<h3>What is acceleration?</h3>

The rate of change of velocity with respect to time is known as acceleration.

Given data;

Initial velocity,u=10m/s

Final velocity,v=40m/s

Time elapsed,t=15 seconds

To find ;

Acceleration, a

The acceleration when the change in the velocity is observed by the formula as:

$\rm a= \frac{v-u}{t}$

Substitute the given values:

$\rm a= \frac{40-10}{15} \\\\ a=\frac{30}{15} \\\\ a=2 \ m/s^2$

The car's acceleration will be 2 m/s².

Hence, option A is correct.

To learn more about acceleration, refer to the link;

brainly.com/question/2437624

#SPJ1

8 0

2 years ago

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance

PSYCHO15rus [73]

Answer:

No she cannot.

Explanation:

Let $v_h$ be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let $v_v$ be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

$t = 95 / v_h$ or $v_h = 95/t$

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

$s = v_vt + gt^2/2= 0$

where g = -9.8m/s2 in the opposite direction with $v_v$

$v_vt - 4.9t^2 = 0$

$v_vt = 4.9t^2$

$v_v = 4.9t$

Since the total velocity that the goal keeper can give the ball is 30m/s

$v = v_v^2 + v_h^2 = 30^2 = 900$

$(4.9t)^2 + \left(\frac{95}{t})^2 = 900$

$24.01t^2 + \frac{9025}{t^2} = 900$

Let substitute x = $t^2$ > 0

$24.01 x + \frac{9025}{x} = 900$

We can multiply both sides by x

$24.01 x^2 + 9025 = 900x$

$24.01x^2 - 900x + 9025 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}$

As $(-900)^2 - 4*24.01*9025 = -56761 < 0$

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.

6 0

3 years ago