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.
If solid, cold metal like brass doesn't melt at room temperature like ice does, than the bass can stand much more heat then a ice cube, so it needs to be hotter to melt, so the melting point of ice is lower then brass's melting point
Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
Answer:
(A) Angular speed 40 rad/sec
Rotation = 50 rad
(b) 37812.5 J
Explanation:
We have given moment of inertia of the wheel 
Initial angular velocity of the wheel 
Angular acceleration 
(a) We know that 
We have given t = 2 sec
So 
Now 
(b) After 3 sec 
We know that kinetic energy is given by 