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Aliun [14]
3 years ago
14

Why is the majority of Earth's freshwater not readily available for our use?

Physics
2 answers:
lara [203]3 years ago
8 0
The answer is B. It is locked up in glaciers and ice caps.

Hope this helped. Good luck! Please give me a thanks and Brainliest.
Simora [160]3 years ago
7 0

Answer:

B.It is locked up in glaciers and ice caps.

Explanation:

Earth is the only planet which we know where water is present is present in all the three forms. Seventy percent of the Earth is filled with water but only 2.5% is freshwater. Rest of it is saline water in oceans. Now, out of this 2.5% freshwater, only 1% is in usable form.  This is because rest of the freshwater is locked up in ice caps and glaciers and cannot be used.

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a 2,000-kilogram railroad car moving at 8m/s to the right collides with a 6,000-kilogram railroad car moving at 3m/s to the west
astra-53 [7]

A freight car of mass 20,000 kg moves along a frictionless level railroad track ... After the push the skateboarder II moves with a velocity of 2 m/s to ... After the collision the cars stick to each other and ... diver jumps with a velocity of 3 m/s in opposite ... A 10 kg object moves at a constant velocity 2 m/s to the right and collides

3 0
3 years ago
A car is on a circular off ramp of an interstate and is traveling at exactly 25 mph around the curve. Does the car have velocity
netineya [11]

Answer:

The car has velocity and acceleration but is not decelerating

Explanation:

Since the car is traveling at 25 mph around the curve, it has a tangential velocity. This tangential velocity is constantly changing in direction (so the car could adapt to the curve and not moving forward in a straight line), there should be a centripetal acceleration in play here. This acceleration does not slow down the car so it's not decelerating.

6 0
3 years ago
A race car, traveling at constant speed, makes one lap around a circular track of radius 200 m. When the car has traveled halfwa
mario62 [17]

The magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

Answer:

Explanation:

Since the race track is a circular track, the distance for one lap will be equal to the circumference of the circular track. And the circumference will be equal to the circumference of the circle.

Since the radius of the track is given as 200 m, then the circumference of the circular track will be

Circumference = 2πr = 2 × 3.14 × 200

So the circumference of the circular track = 1256 m.

So the starting point or position of the track is considered as zero and if the car has traveled half way means, the car has covered half of the circumference of the track.

As the circumference = 1256 m, then half of the circumference of the circle = 1256/2 = 256 m.

So the displacement is the measure of difference between the final position and initial position. As here the initial position is zero and the final position is the halfway around the track which is equal to 256 m.

Then Displacement = Final-Initial = 256-0= 256 m.

So the magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

8 0
3 years ago
Why does a glass rod become positively charged when it is rubbed with a silk cloth?
slavikrds [6]
The core has positive charge<span>, the electrons have negative </span>charge. When you are rubbing<span> the </span>glass rod<span> with the </span>silk cloth<span>, electrons are stripped away from the atoms in the </span>glass<span> and transferred to the </span>silk cloth<span>. This leaves the </span>glass rod<span> with more </span>positive<span> than negative </span>charge<span>, so you get a net </span>positive charge<span>.</span>
5 0
3 years ago
s) A body of mass 2 kilograms moves on a circle of radius 3 meters, making one revolution every 5 seconds. Find the magnitude of
jeyben [28]

Answer:

Centripetal force acting on the body = 9.47 N

Explanation:

Mass of body, m = 2 kg

Radius, r = 3 m

It makes one revolution in 5 seconds.

        Period, T = 5 s

        \texttt{Angular velocity, }\omega =\frac{2\pi }{T}=\frac{2\pi }{5}=1.256rad/s

Centripetal force, F = mrω²

                        F = 2 x 3 x 1.256² = 9.47 N

Centripetal force acting on the body = 9.47 N

8 0
3 years ago
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