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tia_tia [17]
3 years ago
14

Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the cur

rent flowing through the wire, I, and the voltage drop between two points of the wire, V, a known distance apart, L. find the diameter of the wire in terms of \rho L V and I.
How will you find the measurement error (i.e., uncertainty) in the diameter of the wire?
Physics
1 answer:
olganol [36]3 years ago
7 0

Answer:

To find the diameter of the wire, when the following are given:

Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.

Using the ohm's law,

Resistance R = (rho*L)/A

R = V/I.

Crossectional area of the wire A = π*square of radius

Radius = sqrt(A/π)

Diameter = Radius/2 = [sqrt(A/π)]

Making A the subject of the formular

A = (rho* L* I)V.

From the result of A, Diameter can be determined using

Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7

Explanation:

Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

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Examine the images of the Grand Canyon below. Notice that most of the canyon consists of layers of sedimentary rocks, but if you
Tomtit [17]

Intense temperature and pressure of regional metamorphism

Explanation:

The process that cause the formation of the Vishnu Schist is the intense temperature and pressure as a result of regional metamorphism.

  • Regional metamorphism is an extensive metamorphism of an area as a result of temperature and pressure changes.
  • The schist is a foliated metamorphic rock usually found in areas of moderate to high grade temperature and pressure.
  • The Vishnu schist must have been metamorphosed before the new sediments were deposited on top.

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3 years ago
a spaceship, at rest in a certain reference frame s, is given a speed increment of 0.500c. it is then given a further 0.500c inc
bezimeni [28]

In order to give a spaceship at rest in a specific reference frame s a speed increment of 0.500c, seven increments are required. Then, in this new frame, it receives an additional 0.500c increment.

The speed of an object, also known as v in kinematics, is a scalar quantity that refers to the size of the change in that object's position over time or the size of the change in that object's position per unit of time. The distance travelled by an object in a certain period of time divided by the length of the period gives the object's average speed in that period.

The spacecraft moves at v1 = 0.5c after the initial increment.The equation becomes V2 = V+V1/1+V*V1/c after the second one. 2 V2 = 0.5c+0.50c/1+(0.50c)^2/c^ 2 = 0.80c

Likewise, V3 = 0.929c

V4 = 0.976c

V5 = 0.992c

V6 = 0.99c

V7 = 0.999c

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5 0
11 months ago
Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated
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Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN

Explanation: To find the answer we need to know more about the Newton's law of gravitation.

<h3>What is Newton's law of gravitation?</h3>
  • Gravitation is the force of attraction between any two bodies.
  • Every body in the universe attracts every other body with a force.
  • This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
  • Mathematically we can expressed it as,

                        F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2

<h3>How to solve the problem?</h3>
  • Here, we have given with the data's,

                      M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m

  • Thus, the force of attraction between these two bodies will be,

               F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN

Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.

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6 0
1 year ago
a volleyball is hit upward with an initial velocity of 7.5 m/s. calculate the displacement of the volleyball when its final velo
Luden [163]

Answer:

The displacement of the volleyball is 2.62 m

Explanation:

Given;

initial velocity of the volleyball, u = 7.5 m/s

final velocity of the volleyball, v = 2.2 m/s

displacement of the volleyball, d = ?

Apply the following kinematic equation;

v² = u² - 2gd

2gd = u² - v²

d = \frac{u^{2}-v^{2}  }{2g}\\\\d = \frac{7.5^{2}-2.2^{2}  }{2*9.8}\\\\d = 2.62 \ m

Therefore, the displacement of the volleyball is 2.62 m

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