Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 ×
m/s
uniform electric field E = 1.18 ×
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 ×
kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 ×
m/s²
so acceleration is 20.75 ×
m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 ×
= 0 + 20.75 ×
(t)
t = 11.80 ×
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 ×
s
time will elapse before it return to its staring point is 23.6 ns
1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find

2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:

A=(vf-vi)/t
a=(50-25)/10
a=2.5m/s^2
Well, as the waves move it moves the rope as if its trying to take shape of it. Since the rope it light it will move along the ocean and the ocean will keep pushing up on the rope. (even without the waves the water is pushing the rope up so it can take its shape)
Maybe that'll help
<u>Answer:</u>
<em>To relate the type of box material to the warmth inside the box
</em>
<u>Explanation:</u>
3 boxes are made with three different materials glass, plastic and Aluminium. Thermal conductivity is different for different materials. thermal conductors allow easy flow of heat through them and insulators allow minimal or no flow of heat through them.
Thus the amount by which an object gets heated up depends on the value of its conductivity. In this experiment, glass and plastic are insulators and aluminium is a conductor. Among glass and plastic, plastic is a better insulator.
<em>Thus the heat contained in the boxes will be of the order </em>Aluminium>glass>plastic.