A topogeraphical map is diffenent than a physical map in that

An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

$E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta$

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

$\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0$

4 0

3 years ago

You are an astronaut in space far away from any gravitational field, and you throw a rock as hard as you can. The rock will:

Nesterboy [21]

Answer:

the rock will continue at the same speed unless it is affected by another force such as gravity and so if you threw it it will continue to move unless affected by a force

Explanation:

this is because Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

7 0

3 years ago

There are both enviromental and food stimuli that can disrupt sleep true or false

konstantin123 [22]

The answer is true. I hope that this helps you out!!

4 0

2 years ago

Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re

ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of $62 lb / ft ^ 3$ (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

$W = \gamma A * \int_0^15 dy$

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

$W = (62)(14*7)\int^{15}_0 (15-y)dy$