If the ball increases from 1 m/s to 2 m/s, by how much would<br> kinetic energy increase

How to atoms behave in non-magnetic items?

Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

5 0

2 years ago

A container is filled to a depth of 19.0 cm with water. On top of the water floats a 31.0-cm-thick layer of oil with specific gr

IceJOKER [234]

Answer:

Absolute pressure of the oil will be 102822.8 Pa

Explanation:

We have given height h = 31 cm = 0.31 m

Acceleration due to gravity $g=9.8m/sec^2$

Specific gravity of oil = 0.600

So density of oil $\rho =0.6\times 1000=600kg/m^3$

We know that absolute pressure is given by $P=P_0+\rho gh$ , here $P_0=1.01\times 10^5Pa$

So absolute pressure will be equal to $P=1.01\times 10^5+600\times 9.8\times 0.31=102822.8Pa$

So absolute pressure of the oil will be 102822.8 Pa

6 0

3 years ago

What part of the airplane helps to provide it with lift?

Dmitrij [34]

Answer:

Search Results

Featured snippet from the web

The elevators which are on the tail section are used to control the pitch of the plane. A pilot uses a control wheel to raise and lower the elevators, by moving it forward to back ward. Lowering the elevators makes the plane nose go down and allows the plane to go down.

Explanation:

5 0

3 years ago

Categorize the particles according to whether they are found in the nucleus or the electron cloud.

inna [77]

1. neutral particles (neutrons) are in the nucleus
2. nucleus is in the nucleus
3. electron cloud is in the electron cloud
4. positively charged particles (protons) are in the nucleus
5. negatively charged particles (electrons) are in the electron cloud

4 0

3 years ago

Read 2 more answers

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

2 years ago

If the ball increases from 1 m/s to 2 m/s, by how much would kinetic energy increase