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3 years ago

In regards to mechanical energy, is energy being conserved ?

1 answer:

7 0

If only internal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. ... In these situations, the sum of the kinetic and potential energy is everywhere the same.

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Which of the following best explains work?

Anna71 [15]

Answer:

<h2>Which of the following best explains work?</h2>

<h3>Work occurs when a force acts on an object and causes a displacement of that object..</h3>

hope this helps u..

5 0

2 years ago

Read 2 more answers

The baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance, what is th

photoshop1234 [79]

Answer:

9.81 N

Explanation:

Given that a baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance. The maximum height reached will be calculated by using the formula

V^2 = U^2 - 2gH

U = 10 m/s

g = 9.8m/s^2

At maximum height, V = 0

Substitute u and g into the formula

0 = 10^2 - 2 × 9.8 × H

19.6H = 100

H = 100/19.6

H = 5.1 m

The Kinetic energy on the ball will be

K.E = 1/2mv^2

K.E = 1/2 × 1 × 10^2

K.E = 1/2 × 100

K.E = 50 J

But energy = work done

WD = Force × distance (height)

The force that acts on the baseball when it is HALF WAY to the top of the path will be

F × 5.1/2 = 50

F = 100/5.1

F = 19.61 N

The weight acting downward will be

W = mg

W = 1 × 9.8

W = 9.8 N

The net force acting on the ball will be

Net force = F - W

Net force = 19.61 - 9.8

Net force = 9.81 N

Therefore, the net force that acts on the baseball when it is HALF WAY to the top of the path is 9.81 N

4 0

3 years ago

A bowler throws a bowling ball of radius R = 11.0 cm down the lane with initial speed = 8.50 m/s. The ball is thrown in such a w

Ad libitum [116K]

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

Step 1: Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

(a) For what length of time does the ball skid?

The velocity at time t can be written as v(t) = v0 + at

⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

⇒The acceleration (negative) due to friction: a = -µg

⇒ with µ = 0.210

⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t = 46.9*0.11*t = 5.159t

7.219t = 8.5

t = 1.18 seconds

b) How far down the lane does it skid?

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = 8.6 m

c) How many revolutions does it make before it starts to roll?

The angular acceleration of the ball is:

α = τ/I

⇒ with τ = the torque experienced by the ball due the frictional force

⇒ τ = fk*R

α = fk*R /I

⇒ I = 2/5 m*R²

⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

Number of revolutions = 5.19

(d) How fast is it moving when it starts to roll?

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = 6.07 m/s

7 0

3 years ago

Important nuclei of the indirect (multineuronal) system that receive impulses from the equilibrium apparatus of the inner ear an

Lera25 [3.4K]

Answer:

B. Vestibular nuclei

Explanation:

The nerve information generated by the vestibular receptors travels through the vestibular portion of the eighth pair that penetrates the brain stem at the level of the brain stem bridge. At this level there are four vestibular nuclei, which receive the synapses of these axons, coming from the ridges and macules. The semicircular ducts predominantly terminate in the superior and medial nuclei. While the fibers coming from the macules end on the lateral, medial and inferior nuclei. Some fibers of the eighth pair end in the flocculonodular lobe of the cerebellum, these connections play an important role in controlling posture and balance.

From the vestibular nuclei, two bundles of fibers descending to the spinal cord originate from the medial and lateral vestibular spinal bundles and a bundle of fiber that rises in the brain stem that participates in the coordination of eye movements, the medial longitudinal fascicle, which participates in Rotational nystagmus This system also participates in an important way in the control of some ocular movements by the fibers that it contributes to the medial longitudinal fascicle, which is a structure that interconnects the motor nuclei of the extrinsic muscles of the eyeballs VI or abdicens nucleus (abductor) on one side and IV or pathetic nucleus (trochlear) and III or nucleus of the common ocular motor (oculomotor) on the opposite side.

6 0

3 years ago

Identify the 3 types of energy transfer shown in the diagram.

fgiga [73]

Answer:

1.Conduction.

2.Convection.

3.Radiation.

Explanation:

I'm sure it's correct

7 0

3 years ago