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3 years ago

Why is it very difficult for astronomers to find objects like planets and asteroids?

1 answer:

3 0

It is difficult for astronomers to find object like planets and asteroids because it takes a lot of time to verify the objects locations and what surrounds a certain object in order to prove and be precise of its location

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True or false? To develop good alternatives, one should brainstorm ideas and consider different perspectives

Contact [7]

From my experience, I would say it is true.

7 0

3 years ago

A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick

Rom4ik [11]

Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
Decrease in kinetic energy = 4 - 0 = 4 Joules

5 0

2 years ago

What is championship game of baseball called? The World Series The Super Bowl The World Cup

emmasim [6.3K]

Answer:

The World Series

Explanation:

The Super Bowl is the championship American Football game, and the World Cup is the Soccer/Football game.

American Football and Football are different things. The first is what Americans call football, while the other is what Americans call soccer. It is confusing.

6 0

2 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

3 0

2 years ago

2. As a pendulum swings, its energy is constantly converted between kinetic

Alex

Answer:

at point F

Explanation:

To know the point in which the pendulum has the greatest potential energy you can assume that the zero reference of the gravitational energy (it is mandatory to define it) is at the bottom of the pendulum.

Then, when the pendulum reaches it maximum height in its motion the gravitational potential energy is

U = mgh

m: mass of the pendulum

g: gravitational constant

The greatest value is obtained when the pendulum reaches y=h

Furthermore, at this point the pendulum stops to come back in ts motion and then the speed is zero, and so, the kinetic energy (K=1/mv^2=0).

A) answer, at point F

6 0

3 years ago