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Flauer [41]
4 years ago
10

"Pluto Is Changing" talks about sheets of frozen nitrogen on Pluto. Tell what is happening to these sheets of ice

Physics
1 answer:
drek231 [11]4 years ago
7 0
What is Pluto is changing exactly <span />
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There is the more role of the moon than the sun to occur tides in the oceans,why?
Nana76 [90]

Answer:

The ocean tides on earth are caused by both the moon's gravity and the sun's gravity. ... Even though the sun is much more massive and therefore has stronger overall gravity than the moon, the moon is closer to the earth so that its gravitational gradient is stronger than that of the sun.

8 0
3 years ago
The water was used to power a generator, creating _______________ energy
Nikitich [7]

Answer:

electric

<em>please give brainliest</em>

3 0
2 years ago
You should be extra careful during the hours of sunrise, sunset, and nighttime because
skelet666 [1.2K]

Poor visibility, difficulties in colour perception, lessened colour contrast vision due to darker shadows and reduced peripheral vision, these are the reasons why one has to be extra careful while driving during hours of sunrise, sunset and night time.  

<h3><u> Explanation: </u></h3>

Sunrise, sunset and night time are parts of the day with minimal or absolutely no presence of sunlight. To safely navigate roads, we require enough light in order to detect presence of other vehicles, signs and pedestrians. Less sunlight during sunrise and sunset light the sky but makes the roads and vehicles have a darker, less bright view. The contrast between colours is the least, making it difficult to identify objects and see clearly.  

A rising or a setting sun can also lead to glares on the driver’s view and thus obstruct it. Since a change in ambient light is observed, our eyes need to adjust with this change and this isn’t spontaneous. Night time driving has headlight glares from approaching vehicles and reduced surrounding visibility. The eyes switching for vision adaptability from dark to bright light if vehicles approach and pass by is not a quick action. Hence the driver’s vision is compromised in every such case and this may lead to accidents.  

7 0
4 years ago
As a person squeezes and applies pressure to the outside of a balloon, air particles inside
mixas84 [53]

The air particle inside the balloon will collide more with each other and the temperature inside the balloon will increase.

As a person squeezed and applies the pressure to the outside of a balloon, the air particle inside the balloon gains energy and collide with each other, the particle of the air also try leave the balloon surface will implies equal pressure on the wall of the balloon, as the pressure outside the balloon increase, the inside pressure will also increase.

7 0
3 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 2.33 m away from a waterfall 0.488 m in heigh
Ulleksa [173]

Answer:

5.68 m/s

Explanation:

The motion of the salmon is the same as a projectile: it is launched with an initial speed u at an angle of \theta=38^{\circ} above the horizontal.

The motion of the salmon consists of two indipendent motion:

- Along the horizontal direction, it is a uniform motion with constant velocity

v_x = u cos \theta

So that the distance travelled is

d=v_x t = u cos \theta t (1)

- Along the vertical direction, it is a uniformly accelerated motion with constant acceleration downward, so the vertical displacement is

y = u sin \theta t - \frac{1}{2}gt^2 (2)

where g is the acceleration of gravity.

We know the following:

- The horizontal distance travelled by the salmon to reach the waterfall is

d = 2.33 m

- The vertical distance travelled is the height of the waterfall,

y = 0.488 m

From (1) we get:

t=\frac{d}{u cos \theta}

And substituting into (2), we can solve the equation to find t, the time at which the salmon reaches the waterfall:

y = u sin \theta \frac{d}{u cos \theta} - \frac{1}{2}gt^2 = d tan \theta - \frac{1}{2}gt^2\\t = \sqrt{\frac{2(d tan \theta - y)}{g}}=\sqrt{\frac{2(2.33 tan 38^{\circ}-0.488)}{9.81}}=0.521 s

And then, we can use eq.(1) again to find the initial speed, u:

u=\frac{d}{cos \theta t}=\frac{2.33}{cos(38^{\circ})(0.521)}=5.68 m/s

5 0
3 years ago
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