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2 years ago

Do you expect the normal dicrystalline resistance in the forward bias to be pure ohmic resistance or what ?!

Physics

1 answer:

MA_775_DIABLO [31]2 years ago

6 0

Answer:

yes

Explanation:

because is pure ohmic risistance

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Candice is examining a cell under a microscope. She has identified a cell wall, a nucleus, and a chloroplast. What type of organ

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Answer:

A plant

Explanation:

because animals don't have cell walls, and fungus and bacteria dont have chloroplasts

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2 years ago

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2 years ago

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ivolga24 [154]

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2 years ago

A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is

Soloha48 [4]

Answer: $V=5839.051m/s$

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;:

$T=1.26(10)^{4}s$ is the period of the satellite

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=5.98(10)^{24}kg$ is the mass of the Earth

$a$ is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity $V$ is given by:

$V=\sqrt{\frac{GM}{a}}$ (2)

Now, from (1) we can find $a$ , in order to substitute this value in (2):

$a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}$ (3)

$a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}$ (4)

$a=11705845.57m$ (5)

Substituting (5) in (2):

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}$ (6)

$V=5839.051m/s$ (7) This is the speed at which the satellite travels

6 0

2 years ago

) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th

Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

Now, we will substitute 0 for both $z_{1}$ and $z_{2}$ , 0 for w, 334.9 kJ/kg for $h_{1}$ , 2726.5 kJ/kg for $h_{2}$ , 5 m/s for $V_{1}$ and 220 m/s for $V_{2}$ .

Putting the given values into the above formula as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

$1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0$

q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

6 0

3 years ago

Do you expect the normal dicrystalline resistance in the forward bias to be pure ohmic resistance or what ?!​

Do you expect the normal dicrystalline resistance in the forward bias to be pure ohmic resistance or what ?!