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Serga [27]
3 years ago
8

Explain how the atmosphere and hydrosphere interact to create weather

Physics
1 answer:
ELEN [110]3 years ago
3 0

Answer:

The interaction between the atmosphere and hydrosphere usually consist of the water cycle.      

Explanation:

The atmosphere and the hydrosphere interact to create weather activity by the change of states of water, which is usually known as the Water Cycle.

<u>The sun that heats the water from the hydrosphere produces the</u> water evaporation,<u> forming water vapor into the atmosphere</u>. <u>This water vapor is then cooled and</u> <u>condensed </u><u>into the atmosphere</u> to form liquid water, <u>that is released to the hydrosphere as</u> precipitation in form of rain or snow, to star over the cycle.            

I hope it helps you!

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At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta
Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

  • u = initial speed of the automobile = 0 m/s
  • a = constant acceleration of the automobile = 2.5\ m/s^2
  • v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

  • t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0
3 years ago
A rocket is fired at 100 m/s at an angle of 37, how many seconds is it in the air?
Sonbull [250]

Answer:

12.3 seconds

Explanation:

time in air = 2(v)/g

2(100sin(37))/9.8

=12.28 seconds

7 0
3 years ago
It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w
Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0
3 years ago
Use the drop-down menus to complete the statements. When the 5.0 kg cylinder fell 100 m, the final temperature of the water was
OLEGan [10]

Answer:

A. 26.17 B. 1.17 C. 30.86 D. 5.86

Explanation:

7 0
3 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
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