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3 years ago

7

Our family car travels 40 miles in 2 hours. What is our average speed in mph?

1 answer:

skad [1K]3 years ago

3 0

Your average speed is 20 mph, but ONLY IF YOU'RE IN THE CAR during those 2 hours.

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CAN AIR MAKE SHADOWS?

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Answer:

No because you cannot see air so therefore it cannot make shadows

Explanation:

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2 years ago

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Which of the following is a process that occurs at divergent boundaries?

PIT_PIT [208]

The appropriate answer is 4. The formation of Mid-Ocean ridges. There are two types of divergent margins: continental separation and oceanic floor spreading. Where the sea floor separates magma upwells and new oceanic crust is formed. Mid ocean ridges are home to hydrothermal vents. These support a unique ecosystem of marine organisms. The Mid- Atlantic Ridge where the African and Eurasian plates separate from the North and South American plates.

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2 years ago

WHAT IS THE DENSTY OF ALL THE LAYER IN THE ATMOSPHERE

Verdich [7]

Answer:

The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.

Explanation:

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2 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

Zak, helping his mother rearrange the furniture in their living room, moves a 60 kg sofa 5.9 m with a constant force of 40 N. Wh

Len [333]

Answer:236 J

Explanation:

Given

Mass of sofa $m=60 kg$

Displacement $s=5.9 m$

Force $F=40 N$

neglecting Friction

Work done is given by

$W=F\cdot s$

$W=F\cdot s\cos 0$

$W=40\cdot 5.9$

$W=236 J$

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3 years ago

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