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kicyunya [14]
3 years ago
14

What is in the crysphere

Physics
1 answer:
butalik [34]3 years ago
3 0

I think you mean the Cryosphere?

But the answer is D- Earths Ice

This word Cryosphere comes from the greek word "kryos" which means cold

Many people think of the cryosphere as being the north and south poles  but snow and ice can be found in a lot of places on the Earth

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An investigation has been completed similar to the one on latent heat of fusion, where steam is bubbled through a container of w
allochka39001 [22]

Answer:

Q_a=330 cal

Q_w=6000cal

Explanation:

From the question we are told that

Mass of the aluminum container 50 g

Mass of the container and water 250 g

Mass of the water 200 g

Initial temperature of the container and water 20°C

Temperature of the steam 100°C

Final temperature of the container, water, and condensed steam 50°C

Mass of the container, water, and condensed steam 261 g

Mass of the steam 11 g Specific heat of aluminum 0.22 cal/g°C

a) Heat energy on container

Generally the formula for mathematically solving heat gain

      Q_c=M_c *C_c*( \triangle T)

Therefore imputing variables we have

      Q_a=50g *0.22*50-20  

      Q_a=330 cal

b) Heat energy on water

Generally the formula for mathematically solving heat gain

       Q_w=M_w *C_w*( \triangle T)

Therefore imputing variables we have

       Q_w=200 *1* 50-204

       Q_w=6000cal

7 0
3 years ago
Two things you can do to increase the acceleration of an object
stiv31 [10]
You can decrease the mass, or you can increase the force applied to the object
7 0
3 years ago
How are acceleration and speed related​
Gre4nikov [31]

Answer:

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is changing, but not always. When an object moves in a circular path at a constant speed, it is still accelerating, because the direction of its velocity is changing. Comment on robshowsides's post “Speed is the magnitude of velocity.

Explanation:

hope it helped tee hee

6 0
3 years ago
Read 2 more answers
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
4 0
2 years ago
Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat
STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}

Where,

a = Radius

z = Distance to the magnetic field

I = Current

\mu_0 = Permeability constant in free space

Our values are given as

D=2a = 16cm \rightarrow diameter of the sphere then,

a = 0.08m

Thus z = a

B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}

B = \frac{\mu_0I}{2(2^{3/2})a}

B = \frac{\mu_0 I}{2^{5/2}a}

Re-arrange to find I,

I = \frac{2^{5/2}Ba}{\mu_0}

I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}

I = 1.08*10^{-6}A

Therefore the current at the pole of this sphere is 1.08*10^{-6}A

5 0
3 years ago
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