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Blababa [14]
3 years ago
13

What is pollen formation and where does it occur?

Physics
1 answer:
arsen [322]3 years ago
3 0
Pollen formation is the microsporangia of the male cone (flower) and it occurs on the stigma.
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In the diagram, the liquid is vaporizing at which point?
sasho [114]
E is the vapourising state
3 0
4 years ago
PLEASE PLEASE HELP ME!
Vlad1618 [11]
B: heat is transferred as thermal energy by the interaction of moving particles
6 0
3 years ago
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole
ohaa [14]
The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. \alpha is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)

Our final equation is:
g=g_0-g_0(a-1)cos^2(\alpha)
Colatitude is:
\alpha_c=90^\circ-\alpha
The answer is:
g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)

5 0
4 years ago
Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water
tangare [24]

Answer:

T_b=107.3784\ ^{\circ}C

Explanation:

Given:

  • thickness of the base of the kettle, dx=0.52\ cm=5.2\times 10^{-3}\ m
  • radius of the base of the kettle, r=0.12\ m
  • temperature of the top surface of the kettle base, T_t=100^{\circ}C
  • rate of heat transfer through the kettle to boil water, \dot Q=0.409\ kg.min^{-1}
  • We have the latent heat vaporization of water, L=2260\times 10^3\ J.kg^{-1}
  • and thermal conductivity of aluminium, k=240\ W.m^{-1}.K^{-1}

<u>So, the heat rate:</u>

\dot Q=\frac{0.409\times 2260000}{60}

\dot Q=15405.67\ W

<u>From the Fourier's law of conduction we have:</u>

\dot Q=k.A.\frac{dT}{dx}

\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}

where:

A= area of the surface through which conduction occurs

T_b= temperature of the bottom surface

15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}

T_b=107.3784\ ^{\circ}C is the temperature of the bottom of the base surface of the kettle.

6 0
4 years ago
If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net
nika2105 [10]

Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2

6 0
3 years ago
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