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anyanavicka [17]
3 years ago
14

Can radiowaves penetrate clouds​

Physics
1 answer:
lyudmila [28]3 years ago
6 0
Radio waves are reflected, absorbed, scattered, refracted, and diffracted by the atmospheric conditions that they encounter, such as clouds and precipitation. ... For example, only certain wavelengths pass through clouds unimpeded.
You might be interested in
A load of 54 N attached to a spring that ishanging vertically stretches the spring 0.15 m.What is the spring constant?Answer in
beks73 [17]

Answer:

300 N/m

Explanation:

given,

Load attached to the spring, W = 54 N

length of stretch of the spring, x = 0.15 m

spring constant= ?

Force applied on the spring is calculated by the equation

F = k x

where k is the spring constant

x is the displacement of the spring due to applied load

now,

54 = k × 0.15

k = \dfrac{54}{0.15}

k =300\ N/m

hence, the spring constant is equal to 300 N/m

8 0
4 years ago
Read 2 more answers
How much heat is needed to raise the temperature of 50.0 g of water by 25.0°C
love history [14]

Answer:

Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

4.18

J

g

∘

C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

1 g

of that substance by

1

∘

C

.

In water's case, you need to provide

4.18 J

of heat per gram of water to increase its temperature by

1

∘

C

.

What if you wanted to increase the temperature of

1 g

of water by

2

∘

C

? You'd need to provide it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

=

increase by 2

∘

C



2

×

4.18 J

To increase the temperature of

1 g

of water by

n

∘

C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

To cause a

1

∘

C

increase in the temperature of

m

grams of water, you'd need to supply it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

for m g of water



m

×

4.18 J

This means that in order to increase the temperature of

m

grams of water by

n

∘

C

, you need to provide it with

heat

=

m

×

n

×

specific heat

This will account for increasing the temperature of the first gram of the sample by

n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

, and so on until you reach

m

grams of water.

And there you have it. The equation that describes all this will thus be

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q

=

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

50.0

−

25.0

)

∘

C

q

=

10,450 J

Rounded to three sig figs and expressed in kilojoules, t

Explanation:

3 0
3 years ago
Read 2 more answers
A tennis ball is thrown into
liraira [26]

Answer:

Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.

Explanation:

7 0
2 years ago
Please help on this one!!
inysia [295]

The energy transformations that occur as you coast down long hill on a bicycle, including the brakes to make the bike stop at the bottom, is that at the top of the hill you have high GPE AND LOW KE, on your way down you have HIGH KE AND LOW GPE, and at the bottom you have thermal energy due to the stop of the brakes.

the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle and then apply the brakes to make the bike stop at the bottom.


3 0
3 years ago
The current theory of the structure of the
IRISSAK [1]

1) The mass of the continent is 3.3\cdot 10^{21} kg

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

L = 5850 km = 5.85\cdot 10^6 m is the side

t = 35 km = 3.5\cdot 10^4 m is the depth

So the volume is

V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3

We also know that its density is

d=2750 kg/m^3

Therefore, we can find the mass by multiplying volume by density:

m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg

2)

The kinetic energy of the continent is given by:

K=\frac{1}{2}mv^2

where

m=3.3\cdot 10^{21} kg is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is:

v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s

So, we can now find the kinetic energy:

K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

K=\frac{1}{2}mv^2

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0
3 years ago
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