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gtnhenbr [62]
2 years ago
9

Suppose a certain scale is not calibrated correctly, and as a result, the mass of any object is displayed as 0.75 kilogram less

than its actual mass. What is the correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale?
Physics
1 answer:
Papessa [141]2 years ago
7 0

Answer

A certain scale is not calibrated correctly,

Mass displayed  as 0.75 kilogram less than its actual mass.

The scale of display will be equal to '1'

The correlation between actual mass and displayed mass is 1 because correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.

So, the correct answer will be "1"

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Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
2 years ago
A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil
NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ_{crss} = σ_{Y} ( cosФ cosλ)_{max}

now by putting values,

     = (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23

3 0
2 years ago
Four wrestlers step on a scale for a tournament. Wrestler A weighs in at 170 pounds; wrestler B weighs in at 168 pounds; wrestle
OLga [1]
The force due to gravity is equal to the product of the mass and acceleration due to gravity. Since the acceleration due to gravity is constant, we just have to rely on the comparison of their masses to determine the mass effect. Thus, the answer to this item is Wrestler C weighing 171 pounds. 
7 0
2 years ago
Read 2 more answers
PLS ANSWER ASAP!!
Molodets [167]

b) between poles M1 and M2

Explanation:

From the expression, we can deduce that r is the distance between two magnetic poles M1 and M2.

The law of attraction between two magnetic poles states that:

<em>  the force of attraction or repulsion between two magnetic poles is a function of the product of the strength of the magnetic poles and the square of the distance between the pole</em>s

 

    Mathematically:

            FM = K \frac{M1 M2}{r^{2} }

 here r is the distance between the poles

  FM is the magnetic force between the poles

   M1 is the strength of the first magnetic pole

   M2 is the strength of the second pole

   K is the magnetic field constant

learn more:

magnetic pole brainly.com/question/2191993

#learnwithBrainly

8 0
2 years ago
A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field
arsen [322]

magnetic field due to a finite straight conductor is given by

B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)

here since it forms an equilateral triangle so we will have

\theta_1 = \theta_2 = 60 degre

also the perpendicular distance of the point from the wire is

r = \frac{a}{2\sqrt3}

now from the above equation magnetic field due to one wire is given by

B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)

B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)

B = \frac{3\mu_0 i}{2\pi a}

now since in equilateral triangle there are three such wires so net magnetic field will be

B = \frac{9\mu_0 i}{2\pi a}

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