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3 years ago

What is the current flowing through the 20-ohm resistor? A. 1 A

Physics

1 answer:

Rina8888 [55]3 years ago

7 0

Answer:

c a d

Explanation:

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Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stret

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Answer:

29.16 J

Explanation:

From Hook's law,

W = 1/2(ke²)..................... Equation 1

Where W = work done, k = Spring constant, e = extension.

Given: W = 9 J, e = 0.5 m.

Substitute into equation 1

9 = 1/2(k×0.5²)

Solve for k

k = 18/0.5²

k = 72 N/m.

The work done required to stretch the spring by additional 0.4 m is

W = 1/2(72)(0.4+0.5)²

W = 36(0.9²)

W = 29.16 J.

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3 years ago

Most ionic bonds form when electrons from____.

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A bond with elements from B.

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1500 kg, which is required to travel upward 57 m in

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3 years ago

Explain the importance of having a support network when trying to achieve a healthy lifestyle. Who supports you when it comes to

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2 years ago

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A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

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3 years ago