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maks197457 [2]
3 years ago
7

"1. Which properties make a metal a good material to use for electrical wires? (1 poi

Physics
1 answer:
Elanso [62]3 years ago
8 0

Answer:

Malleability

Conductivity

Ductility

<em>Hope</em><em> </em><em>this is</em><em> </em><em>correct and</em><em> </em><em>helpful</em><em> </em>

HAVE A GOOD DAY!

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Which type of electromagnetic radiation cannot be focused?
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The answer is A. Hope this helps. :)
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Study the scenario.
Otrada [13]

If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.

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3 years ago
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If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet
Evgesh-ka [11]

Answer:

S = \dfrac{1}{2.5}

Explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

R = \dfrac{A}{P}

R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}

R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}

R = 4.69 m

using manning's equation

Q = \dfrac{1}{n}AR^{2/3} S^{1/2}

2312= \dfrac{1}{0.013}\times (5(8+5))\times 4.69^{2/3} S^{1/2}

2312=14009.37\times S^{1/2}

S = 0.406

S = \dfrac{1}{2.5}

5 0
3 years ago
El movimiento de los cuerpos y la formación de imágenes son ejemplos de esta clase de fenómeno ​
77julia77 [94]

Answer:

depende de que fenómenos nos referimos de acuerdo al los cuerpos de formación puede aver movimiento contante

7 0
3 years ago
Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express
AlladinOne [14]

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

          Vb = k Q / r        r <R

         

We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

         Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0
3 years ago
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