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3 years ago

Describe an interaction between the cryosphere and the hydrosphere.

1 answer:

3 0

<span>Liquid water exists on the surface in the form of oceans, lakes and rivers on Earth. The frozen part of Earth's hydrosphere is made of glaciers, ice caps and icebergs. The frozen part of the hydrosphere is known as Cryosphere.

In a nutshell:
1) Cryosphere is the frozen part of hydrosphere.
2) Hydrosphere is made up of all bodies of water on the earth.
3) An interaction between Cryosphere and Hydrosphere would be ice caps melting into the ocean.</span>

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How long is a football field?

Nana76 [90]

Answer:

91.44 meters or 300 feet

7 0

3 years ago

Read 2 more answers

A basketball with a mass of 0.5 kilograms is accelerated at 2 meters per second". The nat force acting on the mass is _____

Elanso [62]

Should be d, 1N
hope this helps

7 0

3 years ago

Three point charges are on the x axis: −1 µC

Nina [5.8K]

Answer:

0.0078 N

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

The force is attractive if the two charges have opposite sign, and repulsive if the two charges have same sign.

In this problem, we have:

$q_1 = -1 \mu C = -1 \cdot 10^{-6}C$ located at $x_1=-3 m$

$q_2=+9 \mu C = +9\cdot 10^{-6}C$ located at $x_2=0 m$

$q_3 = -5 \mu C = -5\cdot 10^{-6} C$ located at $x_3=+3 m$

The force between charge 1 and charge 2 is:

$F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.

The force between charge 1 and charge 3 is:

$F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net force on charge 1 is:

$F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N$

towards the right.

5 0

3 years ago

In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

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3 years ago

Rewrite Newton's first law of motion in your own words.

BlackZzzverrR [31]

Newton's first law is called inertia, or the force an object has to resist unbalanced forces that are placed upon it. The more mass an object has, the more inertia an object has. The less mass an object has, the less inertia an object has.

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3 years ago