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Sergio [31]
2 years ago
6

Science help me find the 2 variables please

Physics
2 answers:
dezoksy [38]2 years ago
6 0
I think it's Time and velocity.
victus00 [196]2 years ago
3 0
Time and velocity is the answer
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The circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N. Does the centripetal force do any
Snezhnost [94]

Answer:

Work done is 0.

Explanation:

Given that,

The circumference of an orbit for a toy on a string is 18 m, r = 18 m

Centripetal force, F = 12 N

In the circular path, the centripetal force is always perpendicular to the motion of the object. Thus it makes an angle of 90 degrees with the force and displacement. Hence, we can say that the centripetal force does not do any work on the toy when it follows its orbit for one cycle.

5 0
2 years ago
Read 2 more answers
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
Section 1: Experimental Overview
Luden [163]

Answer:

1-state what the lab is about, that is, what scientific concept (theory, principle, procedure, etc.) you are supposed to be learning about by doing the lab. You should do this briefly, in a sentence or two. If you are having trouble writing the opening sentence of the report, you can try something like: "This laboratory experiment focuses on X…"; "This lab is designed to help students learn about, observe, or investigate, X…." Or begin with a definition of the scientific concept: "X is a theory that…."

2-give the necessary background for the scientific concept by telling what you know about it (the main references you can use are the lab manual, the textbook, lecture notes, and other sources recommended by the lab manual or lab instructor; in more advanced labs you may also be expected to cite the findings of previous scientific studies related to the lab). In relatively simple labs you can do this in a paragraph following the initial statement of the learning context. But in more complex labs, the background may require more paragraphs.

Explanation:

7 0
2 years ago
Solve each of these problems, remembering to include formula, calculations, and units!
KATRIN_1 [288]

Answer:

I nEeD heLp

Explanation:

HelP

5 0
2 years ago
An emf of 28.0 mV is induced in a 501 turn coil when the current is changing at a rate of 12.0 A/s. What is the magnetic flux th
dmitriy555 [2]

Answer:

Φ = 5.589×10⁻⁵  Wb

Explanation:

The inductance of a coil is given as

L = e/(di/dt) ..................... Equation 1

Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.

Also,

The inductance of each turn of the coil when a magnetic field is step up in the coil  is

L = NΦ/i ................. Equation 2

Where N = number of turns, Φ = magnetic field, i = current.

equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

Φ = (e×i)/N.(di/dt) .................. Equation 3

Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

Φ = (0.028×4)/(12×501)

Φ = 0.112/2004

Φ = 5.589×10⁻⁵ Weber

Φ = 5.589×10⁻⁵ Wb

6 0
3 years ago
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