All categories

3 years ago

What is the total precentage of radiation that is reflected by earths atmosphere

1 answer:

6 0

Around 48 % of the incoming solar energy passes through the atmospheric layer of the earth. The water vapor,ozone and dust present in the atmosphere absorbs 23 percent of incoming solar energy .Thus about 71% of the incoming solar radiation is absorbed by the earth. The rest 29% of the incoming solar radiation is reflected back into the space.

You might be interested in

Which type of electromagnetic waves has highest frequency

vesna_86 [32]

Answer:

Gamma rays have the highest energies.

Explanation:

HOPE IT WILL HELP ^_^

6 0

3 years ago

Read 2 more answers

Areas near baton rouge louisiana recently received 36.2 cm of rain in a single day. how many meters of rain was this?

sergey [27]

100 cm is 1 meter. So your answer would be 0.362 meters.

5 0

3 years ago

Read 2 more answers

What is the relationship between the spring constant and the period in a mass hanging on a spring oscillation and why?

anygoal [31]

Explanation:

Period of a mass on a spring is:

T = 2π√(m/k)

T is inversely proportional with the square root of k. So as the spring constant increases, the period decreases.

3 0

4 years ago

Read 2 more answers

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

4 0

3 years ago

When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can

andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

$F=m\frac{v^2}{r}$

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

3 0

3 years ago