Answer:
The radius of the loop is 
Explanation:
Given that,
Current = 0.2kA = 200 A
Number of turns = 35
Magnetic field = 8.9 nT
We need to calculate the radius of one loop
Using formula of magnetic field
Where, I = current
N = number of turns
B = magnetic field
r = radius
Put the value into the formula
Hence, The radius of the loop is
Answer:
The forces of attraction are weak in gases.
Explanation:
Solid-state objects are presented as bodies in a definite form; their atoms are often intertwined into narrowly defined structures, which gives them the ability to withstand forces without apparent deformation. They are generally described as hard as well as resistant, and in them, the forces of attraction are greater than those of repulsion. In the crystalline solids, the presence of small intermolecular spaces gives way to the intervention of the bond forces, which place the cells in geometric forms.
Solid – In a solid, the attractive forces keep the particles together tightly enough so that the particles do not move past each other. Their vibration is related to their kinetic energy. In the solid the particles vibrate in place.
Liquid – In a liquid, particles will flow or glide over one another, but stay toward the bottom of the container. The attractive forces between particles are strong enough to hold a specific volume but not strong enough to keep the molecules sliding over each other.
Gas – In a gas, particles are in continual straight-line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other. In most cases, there are essentially no attractive forces between particles. This means that a gas has nothing to hold a specific shape or volume.
Yes they do observe the law of conservation of mass
Answer:
60 Ω
Explanation:
R(com) = 15 Ω
1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn
1/15 = 1/20 + 1/R2
1/R2 = 1/15 - 1/20
1/R2 = (4 - 3) / 60
1/R2 = 1/60
R2 = 60 Ω
así, la combinada de resistencia necesaria es 60 Ω
The negative charges of the balloon will stick to the positive charges on the wall.
