Answer:
Work = F * s where s is the distance F moves
Since F is stationary, in this case, "no work" is done by either person
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,

- Charge on the second charged particle,

- Position of the first charge =

- Position of the second charge =

The electric field at a point due to a charge
at a point
distance away is given by

where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
.
The electric field at the origin due to first charge is given by

is the position vector of the origin with respect to the position of the first charge.
Assuming,
are the units vectors along x and y axes respectively.

Using these values,

The electric field at the origin due to the second charge is given by

is the position vector of the origin with respect to the position of the second charge.

Using these values,

The net electric field at the origin due to both the charges is given by

Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
<h2>Thus the force of friction is 235 N</h2>
Explanation:
When the bear was at the height of 14 m . Its potential energy = m g h
here m is the mass of bear , g is acceleration due to gravity and h is the height .
Thus P.E = 27 x 10 x 14 = 3780 J
The K.E of the bear just before hitting =
m v²
=
x 27 x ( 6.1 )² = 490 J
The force of friction f = P.E - K.E = 3290 J
Because the work done = Force x Distance
Thus frictional force =
= 235 N
The fragment of an asteroid or any interplanetary material is known as A. METEROID
Answer:
Explanation:
Answer:
Explanation:
Given that,
System of two particle
Ball A has mass
Ma = m
Ball A is moving to the right (positive x axis) with velocity of
Va = 2v •i
Ball B has a mass
Mb = 3m
Ball B is moving to left (negative x axis) with a velocity of
Vb = -v •i
Velocity of centre of mass Vcm?
Velocity of centre of mass can be calculated using
Vcm = 1/M ΣMi•Vi
Where M is sum of mass
M = M1 + M2 + M3 +...
Therefore,
Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb
Rearranging for better understanding
Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)
Vcm = (m•2v + 3m•-v) / (m + 3m)
Vcm = (2mv — 3mv) / 4m
Vcm = —mv / 4m
Vcm = —v / 4
Vcm = —¼V •i