All categories

2 years ago

Please Hurry will mark Brainliest

Physics

2 answers:

Inessa [10]2 years ago

6 0

Answer:

it's not my real number

nikdorinn [45]2 years ago

5 0

Answer:

make me brainliest

Explanation:

ok ok

You might be interested in

Identify each picture as either an inelastic collision or elastic collision

Ivan

Answer:

Inelastic collision:

A collision in which there is a loss of Kinetic Energy due to internal friction of the bodies colliding.

Characteristics of an inelastic collision:

the momentum of the system is conserved
the momentum of the system is conservedloss of kinetic energy

In a perfectly elastic collision, the two bodies that collide with each other stick together.

Elastic collision:

A collision in which the kinetic energy of the two bodies, before and after the collision, remains the same.

Characteristics of elastic collision:

the momentum of the system is conserved
no loss of kinetic energy

In everyday life, no collision is perfectly elastic.

__________________

ANSWER:

Given examples:

Two cars colliding with each other form an example of inelastic collision.

Reason:

(They lose kinetic energy and come to a stop after the collision.)

A ball bouncing after colliding with a surface is an example of elastic collision

Reason:

(a very less amount of kinetic energy is lost)

7 0

2 years ago

According to the concept of punctuated equilibrium, _____.

STatiana [176]

According to the concept of punctuated equilibrium, new species evolve suddenly over relatively short periods of time (a few hundred to a thousand years), followed by longer periods in which little genetic change occurs. Hope this helps. Have a nice day.

7 0

2 years ago

While in a car at 4.47 meters per second a passenger drops a ball from a height of 0.70 meters above the top of a bucket how far

viktelen [127]

Answer:

1.7 m

Explanation:

$v_x$ = Velocity of ball in x direction = 4.47 m/s

$u_y$ = Velocity of ball in y direction = 0

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

t = Time taken

$s_y$ = Vertical displacement = 0.7 m

$s_y=u_yt+\dfrac{1}{2}gt^2\\\Rightarrow 0.7=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{0.7\times 2}{9.81}}\\\Rightarrow t=0.38\ \text{s}$