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3 years ago

Heat from the sun moves through space by the process of

2 answers:

7 0

I don't know what your options are, but heat moves from the sun moves in 3 ways: radiation, conduction, and convection. I hope this helps at least a little

tino4ka555 [31]3 years ago

5 0

Answer: Radiation

Explanation:

There are three modes of heat transfer:

Conduction: Heat transfers from hotter body to colder one via conduction when they are placed in contact with each other. The molecules vibrate and transfer heat.

Convection: Heat transfers in fluids via convection. The heat is carried by bulk motion of fluid.

Radiation: Heat transfer through space takes place via radiation. The electromagnetic waves carry the energy from the Sun and moves through space. This mode does not depend on the medium. The electromagnetic waves carry the energy and travel at the speed of light.

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What are some possible examples of genetic characteristics that can be passed down to child??????

elena-s [515]

They can have a close similar appearance to the parents, have close relation of child reactions.

for example, everyone born in my father's side of the family had the tendency to bump their head on something as they fall asleep up to the point when you are a toddler.

8 0

2 years ago

Read 2 more answers

When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c

e-lub [12.9K]

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = $e^{kt}$

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = $e^{-0.01t}$

V = V₀ $e^{-0.01t}$

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀ $e^{-0.01t}$

0.1V₀ = V₀ $e^{-0.01t}$

0.1V₀/V₀ = $e^{-0.01t}$

0.1 = $e^{-0.01t}$

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0

2 years ago

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453

Drupady [299]

Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

$0.0001\times \frac{100}{0.4539}=0.022%$

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

$\frac{1}{0.022}\times {100}=4545.5\ kg$

Mass in pounds would be

$\frac{4545.5}{0.4539}=10014.32\ lb$

Mass in pound-mass is 10014.32 lb

8 0

2 years ago

Mercury, Venus, Earth, and Mars are all ______ planets.

Pepsi [2]

Mercury, Venus, Earth, and Mars are all ___inner___ planets. This is because they are all within the asteroid belt.

8 0

2 years ago

A ball is thrown downward at 12 m/s from a windowsill 35 m above the ground. At the same time, another ball is thrown upward at

wariber [46]

Answer:

The second ball lands 1.5 s after the first ball.

Explanation:

Given;

initial velocity of the ball, u = 12 m/s

height of fall, h = 35 m

initial velocity of the second, v = 12 m/s

Time taken for the first ball to land;

$t = \sqrt{\frac{2h}{g} }\\\\t =\sqrt{ \frac{2*35}{9.8}}\\\\t = 2.67 \ s$

determine the maximum height reached by the second ball;

v² = u² -2gh

at maximum height, the final velocity, v = 0

0 = 12² - (2 x 9.8)h

19.6h = 144

h = 144 / 19.6

h = 7.35 m

time to reach this height;

$t_1 = \sqrt{\frac{2h}{g} }\\\\t_1 = \sqrt{\frac{2*7.35}{9.8}}\\\\t_1 = 1.23 \ s$

Total height above the ground to be traveled by the second ball is given as;

= 7.35 m + 35m

= 42.35 m

Time taken for the second ball to fall from this height;

$t_2 = \sqrt{\frac{2h}{g} }\\\\t_2 = \sqrt{\frac{2*42.35}{9.8} }\\\\t_2 = 2.94 \ s$

total time spent in air by the second ball;

T = t₁ + t₂

T = 1.23 s + 2.94 s

T = 4.17 s

Time taken for the second ball to land after the first ball is given by;

t = 4.17 s - 2.67 s

T = 1.5 s

Therefore, the second ball lands 1.5 s after the first ball.

4 0

2 years ago