The compass doesn’t give you the value of the net magnetic field, just the direction. So, how do you get the magnitude of a particular field from this? The trick is to assume the value of the Earth’s magnetic field and the direction of the compass. Let’s assume that at this location on the Earth, the magnetic field is pointing directly North with a horizontal component of about 2 x 10-5 T.
Now suppose that I do something to create a magnetic field in a known direction and perpendicular to the horizontal component of the Earth’s magnetic field. Here is an example where I put a current carrying wire right over the compass needle. Since the compass is underneath the wire, the magnetic field due to the wire will be 90° to the Earth’s magnetic field.
The positively charged atmosphere attracts negatively charged spider silk, might electrostatic force play in spider dispersal, according to a recent study.
Answer: Option C
<u>Explanation:</u>
The positive charge present in upper of the atmosphere and the negative charge on planet’s surface. During cloudless skies days, the air possesses a voltage of nearly around 100 volts for each and every meter from above the ground.
Ballooning spiders process within this planetary electric field. When their silk relieve their bodies then it picks up a negative charge. This oppose the similar negative charges on the surfaces on which the spiders settles and create sufficient force to lift them into the air. And spiders can hike those forces by climbing onto blades of grass,twigs, or leaves.
Answer:
Explanation:
The x component is the adjacent side making up the given angle (39.4)
The vector is the hypotenuse.
The definition of the cos (x) is adjacent / hypotenuse.
cos(39.4) = adjacent / 47.3 Multiply both sides by 47.3
47.3 * cos(39.4) = adjacent Cos(39.4) = 0.7727
adjacent = 36.55
Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
