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3 years ago

9

Which of these processes describes the effect Earth's atmosphere has on Earth's hydrosphere?

1 answer:

Ede4ka [16]3 years ago

6 0

Warm, moist air increasing ocean temp

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Can someone answer my last question you could get 25 points total

enyata [817]

Answer:

Hi where is question. Hope you understand me

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2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

7 0

3 years ago

IF YOU ANSWER THESE 4 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!

melisa1 [442]

Answer:

1.d

2.d

3.c

4.c

Okay, d & d, c & c. Iknow what you're thinking, I guessed. Nope! I just took a good look and chosed what I thought was the best answer

Explanation:

Have a great day! Byeeee

4 0

3 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path, it has a velocity of 34.5

Inessa05 [86]

Answer:

the acceleration of the airplane is 5.06 x 10⁻³ m/s²

Explanation:

Given;

initial velocity of the airplane. u = 34.5 m/s

distance traveled by the airplane, s = 46,100 m

final velocity of the airplane, v = 40.7 m/s

The acceleration of the airplane is calculated from the following kinematic equation;

v² = u² + 2as

$2as= v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(40.7)^2 -(34.5)^2}{2 \times 46,100} \\\\a = 5.06 \ \times \ 10^{-3} \ m/s^2$

Therefore, the acceleration of the airplane is 5.06 x 10⁻³ m/s²

5 0

3 years ago

If the box weighs 40 newtons and is lifted a distance of 2 meters, how much work is done on the box?

Zarrin [17]

Answer:

The work done on the box is 80 J.

Explanation:

Given that,

Weight of box = 40 N

Distance = 2 meter

We need to calculate the work done

Using formula of work done

$W=F\times x$

$W=mg\times x$

Where, x = distance

mg = weight

Put the value into the formula

$W=40\times2$

$W= 80\ Nm$

$W=80\ J$

Hence, The work done on the box is 80 J.

5 0

3 years ago