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1 year ago

Which planet might best be described as a large dirty ice ball?

Physics

2 answers:

Dimas [21]1 year ago

4 0

The planet that is BEST described as a dirty ice ball would be the planet Pluto

hope this helps

Diano4ka-milaya [45]1 year ago

3 0

This planet would be known as pluto

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Lorenzo is making a prediction. “I learned that nonmetals increase in reactivity when moving from left to right. So I predict th

nadezda [96]

That prediction is not correct because Xenon is extremely stable; column 18 of the periodic table contains the noble gasses, which are stable because their outer-most energy levels are completely filled. Having the octet (8) of valence electrons means that the element no longer needs to lose or gain electrons to gain stability.

The column 17 elements are unstable because they only have one valence electron short of the stable octet configuration of the noble gasses.

6 0

2 years ago

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A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the distance the car travels during this time.

andrew11 [14]

It traveled 39.6 meters

6 0

2 years ago

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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago

A food department is kept at â12Â°c by a refrigerator in an environment at 30Â°c. the total heat gain to the food department is

slava [35]

As per energy conservation in the reversible engine we can say

$Q_2 + W = Q_1$

here we know that

$Q_2 = 3300 kJ/h$

$Q_1 = 4800 kJ/h$

now from above equation

$3300 + W = 4800$

$W = 1500 kJ/h$

now we can convert it into kW

$W = 1500\times \frac{kJ}{3600s}$

$W = 0.42 kW$

so above is the power input to the refrigerator

now to find COP we know that

$COP = \frac{Q_2}{W}$

$COP = \frac{3300}{1500} = 2.2$

so COP of refrigerator is 2.2

3 0

1 year ago

The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision betwee

enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

As the collision is perfectly elastic, kinetic energy must be conserved.
The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

$v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}$

As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

The maximum energy stored in the in the spring is given by the following expression:

$U =\frac{1}{2} *k * A^{2}$

where A = maximum compression of the spring.

This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
As total kinetic energy must be conserved, the following condition must be met:

$KE_{10} = KE_{1f} + KE_{2f}$

So, it is clear that KE₂f < KE₁₀
Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

As explained above, if total kinetic energy must be conserved:

$KE_{10} = KE_{1f} + KE_{2f}$

So as kinetic energy is always positive, KEf₂ < KE₁₀.

4 0

1 year ago