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lianna [129]
3 years ago
9

Present day glaciers are found primarily in _______________.

Physics
2 answers:
Mazyrski [523]3 years ago
8 0
<h3><u>Answers;</u></h3>

Antarctica and Greenland

Present day glaciers are found primarily in <em><u>Antarctica and Greenland</u></em>.

<h3><u>Explanation;</u></h3>
  • <em><u>The two major ice sheets that exists today are found primarily in Antarctica and Greenland. Ice sheets are large masses of glacial ice that are also known as continental glaciers.</u></em>
  • Most ice in Antarctica and Greenland spill out into the ocean from a few spots. The Antarctica and Greenland ice sheets combined comprise more than 99 percent of freshwater ice found on Earth.
KonstantinChe [14]3 years ago
4 0
<span>Antarctica and Greenland

</span>
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a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th
Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

  Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>

<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock.  A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

    His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock.  The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so 

                     (300 x speed) = 36 kg-m/s .

Divide each side by 300:  speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum.  The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed 
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
                                                         (300 x speed) = 186

Divide each side by 300:  speed = 186/300 = <em>0.62 m/s</em>    <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.

  yay !

By the way ... thanks for the 6 points.  The warm cloudy water
and crusty green bread are delicious.


4 0
3 years ago
Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r
Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

v=u-gt

u=gt

t=\dfrac{u}{g}

t=\dfrac{13.7}{9.8}

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

h=ut+\dfrac{1}{2}at^2

Here, a = -g

h=ut-\dfrac{1}{2}gt^2

h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0
3 years ago
temperatura en la ciudad de la página madrugada y 2 grados bajo cero y a las 8 de la mañana sube a 6 grados y luego ya a las 10
vivado [14]

Answer:

kjkjnklhklsehfiodxlfngioduherl

Explanation:

8 0
2 years ago
Read 2 more answers
Which type of physical activity is being performed in the picture?
mafiozo [28]
B strength training I think that’s the answer
3 0
3 years ago
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a watermelon is balanced by a dog, a pumpkin, a flowerpot, and a baseball as shown below. Is the weight of the watermelon equal
alukav5142 [94]

Answer:

pumpkin

Explanation:

watermelon and pumpkins are close to shape and size

6 0
3 years ago
Read 2 more answers
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