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3 years ago

Scientists are expected to share their results by

Physics

1 answer:

shepuryov [24]3 years ago

7 0

Answer:

Publishing results of research projects in peer-reviewed journals enables the scientific and medical community to evaluate the findings themselves. It also provides instructions so that other researchers can repeat the experiment or build on it to verify and confirm the results.

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Thalia is drafting a plan to move a large, perfect sphere concrete sculpture that is in front of her office building. Describe t

Marina86 [1]

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Thalia must consider the weight of the object and the mass of the sculpture. Weight and mass are different things. She should also consider the time on how long it will take to move it and where she'll move it.

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3 years ago

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Current is produced when charges are accelerated by an electric field to move to a position of lower

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Current is created when charges are quickened by an electric field to move where the position of lower temperature. An electric current is a stream of electric charge. In electric circuits, this charge is regularly conveyed by moving electrons in a wire.

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3 years ago

What is the name of the hair-like structures on sponge cells that move back and forth to help move water, nutrients, and waste t

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3 years ago

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Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

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3 years ago

At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th

Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

8 0

3 years ago