All categories

2 years ago

A 0.15-kg ball is thrown into the air and rises to a height of 20.0 m. How much kinetic energy did the ball initially have?

2 answers:

7 0

IF the toss was straight upward, then the kinetic energy it got
from the toss is the gravitational potential energy it has at the top,
where it stops rising and starts falling.

Potential energy = (mass) x   (gravity) x (height)

   = (0.15 kg) x (9.8 m/s²) x (20 m)

   = 29.4 kg-m²/s² = 29.4 joules .

mel-nik [20]2 years ago

3 0

The conservation of energy says that energy in an isolated system remains constant, that it is not created nor destroyed but transforms from one form of energy to another.
So when the ball is thrown into the air, it has kinetic energy (KE). As it goes into the air, that kinetic energy transforms into gravitational potential energy (GPE). The higher it gets, the more GPE it has and therefore the less KE it has.
At the top, all of it is transformed into GPE and there is no KE.
But total energy at any point after the ball is thrown until it lands it has the same energy throughout.

Knowing this, you can set the beginning KE equal to the final GPE.
KE = GPE

Insert formula for GPE.

KE = mgh
KE = 0.15kg*9.8m/s^2*20.0m
KE = 29 J

The ball initially had 29 J of kinetic energy

You might be interested in

Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re

lapo4ka [179]

Answer:

$K_{system} = \frac{k}{10}$

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})$

Here, n = 10

Spring constant of each spring = k

Thus,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})$

$\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})$

$\frac{1}{K_{system}} = \frac{10}{k}$

$K_{system} = \frac{k}{10}$

7 0

3 years ago

Read 2 more answers

What is the mass of an object that requires 100N of force in order to accelerate 10m/s2

iragen [17]

Explanation:

Hey there!

Given;

Force (f) = 100 N

Acceleration (a) = 10m/s^2.

Mass (m)=?

We have;

Force = mass * acceleration.

So,

100 = m * 10

Or, 100 = 10m

Or, m= 100/10

Therefore, mass is 10kg.

Hope it helps....

3 0

2 years ago

What properties are those that describe what happens when a substance reacts with another substance?

Vanyuwa [196]

Chemical properties due to the changes in both

6 0

3 years ago

Read 2 more answers

When 2 grams of hydrogen (h2) react with 32 grams of oxygen (O2) the reaction yields hydrogen peroxide (h2o2) how many grams of

MissTica

as per the reaction we have

$H_2 + O_2 ---> H_2O_2$

so here 1 mole of H2 react with 1 mole of O2 and give 1 mole of H2O2 as the product

now we will find the moles of given reactant

moles of H2

$n_1 = \frac{2}{2} = 1 mole$

moles of O2

$n_2 = \frac{32}{32} = 1 mole$

so here both present in 1 mole each hence the product will also formed 1 mole

so here we have

$n = 1 mole$ H2O2 will form

so mass of the product

mass = moles* molecular mass

$m = 1 * (2 + 32)$

$m = 34 g$

so the product mass will be 34 g

6 0

2 years ago

Which of the phases below are able to be compressed? A) Liquid only. B) Liquid and Gas. C) Solid only. D) Gas only

amid [387]

The phrase below that is able to be compressed is D. gas only.
Although it is possible to compress liquids, it requires a great deal of pressure to accomplish a little compression, which is why liquids and solids are often referred to as incompressible.

3 0

2 years ago

Read 2 more answers