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3 years ago

Compare the densities of two objects that have the same volume, but one feels heavier than the other.

1 answer:

8 0

We know, density = Mass / Volume
It represents density is directly proportional to mass and indirectly proportional to volume. So, at a constant volume, object with larger density will appear more heavier than that of object with smaller density.

In short, Heavier object will have the higher density than the other.

Hope this helps!

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A wire runs left to right and carries a current in the direction shown.

Step2247 [10]

Answer:

The direction of the magnetic field at point Z; Into the screen

Explanation:

8 0

3 years ago

Read 2 more answers

A person kicks a 4.0-kilogram door with a 48-newton force causing the door to accelerate at 12 meters per second squared. What i

inna [77]

Answer:

-48 N

Explanation:

mass of door (m) = 4 kg

acceleration of the door = 12 m/s^{2}

force exerted by the person = 48 N

From Newton's third law of motion, action and reaction are equal but opposite. Therefore the force exerted on the door by the person which is 48 N will be the same as the force exerted on the person by the door but opposite in its direction, and this would be - 48 N

7 0

3 years ago

What si the name of the state change water is going through as it boils?

Blababa [14]

It evaporates into a vapor

~~~hope this helps~~~
~~have a beautiful day~~
~davatar~

3 0

3 years ago

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.

8 0

4 years ago

What is the formula of finding displacement

xenn [34]

Displacement is the final position of the object minus the initial position of the object.
Xf - Xi. Displacement is not the distance of the object. If you go to the right 10m and to the left another 10m, your displacement is 0m. But your distance is 20m

8 0

4 years ago