2 coplas o pregones inventadas por ti relacionadas con la región caribe.

Describe the relationship between kinetic energy and speed and give an example of how changing and object speed would affect its

Andreas93 [3]

Answer:

I'm not sure

Explanation:

I have had that question to Uchida c r go crew in to go be

6 0

2 years ago

A woman pushes a 27 kg lawnmower at a steady speed. She exerts a 120 N force in a direction 35◦ below the horizontal. The accele

nikklg [1K]

The vertical force exerted on the lawn is 68.8 N downward

Explanation:

The vertical force exerted by the lawnmower on the lawn is equal to the vertical component of the force applied, therefore:

$F_y = F sin \theta$

where

F is the magnitude of the force applied

$\theta$ is the angle between the direction of the force and the horizontal

In this problem:

F = 120 N

$\theta=-35^{\circ}$

Substituting,

$F_y = (120)(sin 30)=-68.8 N$

where the negative sign means the direction of the force is downward.

Learn more about vector components and forces here:

brainly.com/question/2678571

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

3 0

3 years ago

a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

3 years ago

Estimate the length of the neptunian year using the fact that the earth is 1.50Ã108km from the sun on average.

Ksenya-84 [330]

Answer:

Explanation:

We know that,

Neptune is 4.5×10^9 km from the sun

And given that,

Earth is 1.5×10^8km from sun

Then,

Let P be the orbital period and

Let a be the semi-major axis

Using Keplers third law

Then, the relation between the orbital period and the semi major axis is

P² ∝ a³

Then,

P² = ka³

P²/a³ = k

So,

P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³

Period of earth P(earth) =1year

Semi major axis of earth is

a(earth) = 1.5×10^8km

The semi major axis of Neptune is

a (Neptune) = 4.5×10^9km

So,

P(E)²/a(E)³ = P(N)² / a(N)³

1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³

Cross multiply

P(N)² = (4.5×10^9)³ / (1.5×10^8)³

P(N)² = 27000

P(N) =√27000

P(N) = 164.32years

The period of Neptune is 164.32years

6 0

3 years ago

There is a plate with moment of inertia Ip = 0.0711 kgm2 , rotating around its center of mass with angular speed of 4.53 rad/s.

Anna007 [38]

Answer:

1) their common angular speed = 3.038 rad/s

2) kinetic energy loss = 0.24J

Explanation:

1) This is a case of conservation of angular momentum.

The initial angular momentum must be equal to the final angular momentum

Initial angular momentum = I x w

Where I = moment of inertia, and

w = angular momentum.

Initial angular momentum = 0.0711 x 4.53 = 0.322 kg-m2-rad/s

After addition of ring, moment of inertia becomes,

0.0711 + 0.0353 = 0.106 kg-m2

Therefore, final angular momentum will be

0.106 x Wf

Where Wf = final common angular velocity

Equating the two angular momentum we have

0.322 = 0.106Wf

Wf = 0.322/0.106 = 3.038 rad/s

2) KE = 1/2 x I x w^2

Initial KE = 1/2 x 0.0711 x 4.53^2

= 0.729 J

Final KE = 1/2 x 0.106 x 3.038^2

= 0.489 J

Loss in KE = 0.729 - 0.489 = 0.24 J

5 0

3 years ago