All categories

2 years ago

Clouds absorb outgoing radiation emitted by earth and reradiate a portion of it back to the surface during _____.

1 answer:

4 0

I would say the correct answer is nighttime. The clouds in the atmosphere would absorb the radiation that is emitted by the Earth and at at night 30% of these would be radiated back to the Earth's surface. Hope this answers the question. <span />

You might be interested in

PLEASE HELP ASAP!!!!!!!!!!WILL GIVE BRAINLIEST...

Delicious77 [7]

The answer to this is D. Green.

7 0

2 years ago

Read 2 more answers

A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

galben [10]

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

5 0

2 years ago

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in

Lana71 [14]

Answer:

The average power is calculated as 735.0 W

Solution:

As per the question:

Total mass, M = 1200 kg

Counter mass of the elevator, m = 950

Distance traveled by the elevator, d = 54 m

Time taken, t = 3 min = 180 s

Now,

To calculate the average power:

First, we find the force needed for lifting the weight:

Force, F = (M - m)g = $(1200 - 950)\times 9.8 = 2450 N$

Now, the work done by this force:

W = Fd = $2450\times 54 = 132300\ J = 132.3\ kJ$

Average power is given as:

$P_{avg} = \frac{W}{t} = \frac{132300}{180} = 735.0\ W$

4 0

2 years ago

The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which

spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒ $\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

On putting estimate values, we get

⇒ $\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}$

⇒ $\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}$

⇒ $v=1.94 \ cm$

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

$m=\frac{v}{u}$ and $m=\frac{h_{1}}{h_{0}}$

⇒ $\frac{v}{u} =\frac{h_{1}}{h_{0}}$

⇒ $\frac{1.94}{25}=\frac{{h_{1}}}{15}$

⇒ ${h_{1}}=1.16 \ mm$

8 0

3 years ago

A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N

irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

3 0

2 years ago