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3 years ago

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How is the moon's position in a total solar eclipse different from its position in an annular eclipse? Please write in your own

words

1 answer:

Savatey [412]3 years ago

4 0

Answer:

The difference of the moons position when a solar eclipse happens is it is closer to earth when a annular eclipse happens it is farther away from earth.

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

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4 years ago

URGENT: write a simple equation relating the frequency (f) of a wave to its period (T)

EastWind [94]

Answer:

f(x) = mx + b

Explanation:

5 0

3 years ago

The amount of energy invested into glycolysis is ____________ atp

Liono4ka [1.6K]

The amount of energy invested into glycolysis is 2 ATP.

3 0

3 years ago

THE ONE WHO WILL ANSWER IT CORRECT WILL BE MARKED AS "BRAINLIST"....... If the radius of the orbit of a spaceship from the centr

jonny [76]

Answer:

Explanation:

19116 being the distance from the centre of the earth to the point you want to check in metres. So after all that you get G = 1.0888m/second squared. Hope that helps you/ you should be able to pop in any height you want.

plz mark as brainliest

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3 years ago

MARKING BRAINLIEST! The velocity of sound waves in air is 343 m/s, 1,007 m/s in helium and 267 m/s in carbon dioxide. Which one

FromTheMoon [43]

Answer:

In helium

Explanation:

Since the units are the same, the value should only be considered.

6 0

3 years ago

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