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3 years ago

Place the lunar phases in the correct order

1 answer:

tatiyna3 years ago

4 0

Day zero. . . . . . . . . . New Moon
First 7.4 days . . . . . . Waxing Crescent
End of 7.4 days. . . . . First Quarter
Next 7.4 days . . . . . . Waxing Gibbous
End of 14.8 days. . . . Full Moon
Next 7.4 days . . . . . . Waning Gibbous
End of 22.1 days. . . . Third Quarter
Next 7.4 days. . . . . . . Waning Crescent
End of 29.53 days . . . Next New Moon

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Which scenario did not include a chemical change?

Jobisdone [24]

Answer:

what scenario i dont understand

Explanation:

step by step explenation

7 0

3 years ago

For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m

Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is $6.85\times 10^{5}$ newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

$\Sigma F = F-R = 0$ (Eq. 1)

Where:

$F$ - Forward trust, measured in newtons.

$R$ - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: ( $F = 6.85\times 10^{5}\,N$ )

$R = F$

$R = 6.85\times 10^{5}\,N$

The magnitude of the resistive force exerted by the water is $6.85\times 10^{5}$ newtons.

4 0

4 years ago

The images show two different types of Galapagos tortoises that scientists believe descended from the same species. The first ty

vichka [17]

Answer:

C. turtles with genes for long necks had a better chance of surviving to reach reproductive age.

Explanation:

The turtles that had long necks were more fit to the environmnet in which they were lovated and were able to grow larger and have more reproductive time because of their ability to feed on grass and small shrubs, this helped them always haev food available, and made them the dominant gene eventually.

3 0

4 years ago

Read 2 more answers

What is the best use of an atomic model to explain the charge of the particles in thomson's beams

hoa [83]

The best use of an atomic model to explain the charge of the particles in Thomson's beams is:

<u>An atom's smaller negative particles are at a distance from the central positive particles, so the negative particles are easier to remove.</u>

<u>Explanation:</u>

In Thomson's model, an atom comprises of electrons that are surrounded by a group of positive particles to equal the electron's negative particles, like negatively charged “plums” that are surrounded by positively charged “pudding”.

Atoms are composed of a nucleus that consists of protons and neutrons . Electron was discovered by Sir J.J.Thomson. Atoms are neutral overall, therefore in Thomson’s ‘plum pudding model’:

atoms are spheres of positive charge
electrons are dotted around inside

Thomson's conclusions made him to propose the Rutherford model of the atom where the atom had a concentrated nucleus of positive charge and also large mass.

6 0

3 years ago

At a given instant an object has an angular velocity. It also has an angular acceleration due to torques that are present. There

katen-ka-za [31]

a) Constant

b) Constant

Explanation:

We can answer this question by using the equivalent of Newton's second law of motion of rotational motion, which can be written as:

$\tau_{net} = I \alpha$ (1)

where

$\tau_{net}$ is the net torque acting on the object in rotation

I is the moment of inertia of the object

$\alpha$ is the angular acceleration

The angular acceleration is the rate of change of the angular velocity, so it can be written as

$\alpha = \frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time interval

So we can rewrite eq.(1) as

$\tau_{net}=I\frac{\Delta \omega}{\Delta t}$

In this problem, we are told that at a given instant, the object has an angular acceleration due to the presence of torques, so there is a non-zero change in angular velocity.

Then, additional torques are applied, so that the net torque suddenly equal to zero, so:

$\tau_{net}=0$

From the previous equation, this implies that

$\Delta \omega =0$

Which means that the angular velocity at that instant does not change anymore.

In this second case instead, all the torques are suddenly removed.

This also means that the net torque becomes zero as well:

$\tau_{net}=0$

Therefore, this means that

$\Delta \omega =0$

So also in this case, there is no change in angular velocity: this means that the angular velocity of the object will remain constant.

So cases (a) and (b) are basically the same situation, as the net torque is zero in both cases, so the object acts in the same way.

8 0

4 years ago