All categories

3 years ago

Which area of earth is most similar to the suns convention zone

1 answer:

3 0

The area of the Earth that is most similar to the Sun's convection zone would be the mantle. The convection zone of the sun is its outermost layer where heat transfer by convection happens which is similar to the Earth's mantle. It would be the crust because it is the outer most layer.

You might be interested in

A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

3 years ago

Couldn’t you technically make infinite speed by putting a car in a vacuum chamber? Since top speed it made by the amount of forc

kakasveta [241]

Answer:

Explanation:

For infinite speed to be achevied, one must have no sink of energy to spend. The source of entropy in this example, is the tires hitting the surface, producing heat and friction. Not to mention that you'd still need fuel to start the car, and an infinite tunnel or track, which would be impossible and speed up to process of energy loss through entropy quicker.

7 0

3 years ago

You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of po

nataly862011 [7]

Answer:

a) q = 39.29 cm , b) h ’= - 3.929 cm the image is inverted and REAL

Explanation:

For this exercise we will use the equation of the constructor

1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

f = R / 2

f = 44/2

f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

1 / q = 1 / f - 1 / p

1 / q = 1/22 - 1/50

1 / q = 0.0254

q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

m = h’/ h = - q / p

h ’= - q / p h

h ’= - 39.29 / 50 5

h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

4 0

3 years ago

A particle with a charge of 5.13 μC has a velocity of 8.64 x106 m/s in a direction perpendicular to a magnetic fieldof 1.99 x 10

Svetach [21]

Answer:

F= 0.009 N

Explanation:

Given that

Charge ,q= 5.13 μC

Velocity ,V= 8.64 x 10⁶ m/s

Magnetic field , B = 1.99 x 10⁻⁴ T

The force on a charge q moving with velocity v is given as follows

F= q V B

Now by putting the values in the above equation we get

[tex]F= 5.13\times 10^{-6}\times 8.64\times 10^{6}\times 1.99\times 10^{-4}\ N [\tex]

F=0.00882 N

F= 0.009 N

Therefore the force on the particle will be 0.009 N.

5 0

3 years ago

The Moon (7.35 x 10^22 kg) is 3.85 x 10^8 m from the Earth and orbits around it in a circle. a) If the period of the Moon is 27.

Lemur [1.5K]

a) Speed of the Moon: 1025 m/s

The speed of the moon is equal to the ratio between the circumference of its orbit and the orbital period:

$v=\frac{2 \pi r}{T}$

where

$r=3.85\cdot 10^8 m$ is the radius of the orbit of the Moon

$T=27.3 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=2.36\cdot 10^6 s$ is the orbital period

Substituting into the formula, we find

$v=\frac{2 \pi (3.85\cdot 10^8 m)}{2.36\cdot 10^6 s)}=1025 m/s$

b) Centripetal force: $2.0 \cdot 10^{20} N$

The centripetal force acting on the Moon is given by:

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

where

$m=7.35 \cdot 10^{22}kg$ is the mass of the Moon

$v=1025 m/s$ is its orbital speed

$r=3.85\cdot 10^8 m$ is the radius of the orbit

Substituting into the formula, we find

$F=(7.35 \cdot 10^{22} kg) \frac{(1025 m/s)^2}{3.85\cdot 10^8 m}=2.0 \cdot 10^{20} N$

c) Gravitational force

The only relevant force that acts on the Moon, and that keeps the Moon in circular motion around the Earth, is the gravitational force exerted by the Earth on the Moon. In fact, this force "pulls" the Moon towards the Earth, so towards the centre of the orbit of the Moon, therefore it acts as source of centripetal force for the Moon.

8 0

3 years ago