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3 years ago

Hurry help please, number 3

Physics

2 answers:

STatiana [176]3 years ago

8 0

C) Most of earths weather occurs in this layer.

Dovator [93]3 years ago

6 0

C most of the Earths weather occurs in this layer.

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Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo

luda_lava [24]

Answer:

Explanation:

8 0

2 years ago

A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?

Genrish500 [490]

A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps

8 0

3 years ago

Read 2 more answers

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

3 years ago

A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:

mario62 [17]

Resistance=R=10ohm
Voltage=V=5V
Current=I

Applying ohm's law

$\\ \sf\longmapsto \dfrac{V}{I}=R$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{5}{10}$

$\\ \sf\longmapsto I=0.5A$

4 0

2 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:

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:

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

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:

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

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