Answer:
the force acting on the team mate is 1.19 kN.
Explanation:
given,
mass = 196 lbm
while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s
time taken for deceleration = 0.5 sec
F = mass × acceleration
acceleration =
= -13.4 m/s²
1 lbs = 0.453 kg
196 lbs = 196 × 0.453 = 88.79 kg
F = 88.79 × 13.4
F = 1189.786 N = 1.19 kN
hence, the force acting on the team mate is 1.19 kN.
Answer:
Explanation:
The incident angle is so the reflected ray will also be
.
The normal divides the angle between the incident and reflected ray. So, the angle between them would be
So, the angle between the incident and reflected rays is .
a) Speed: It is the ratio of distance covered and time interval to cover that distance
it is given as
b) Direction: it will define for vector quantities and shows the required direction for it
c) Velocity : It is rate of change in position or it is ratio of displacement and time
d) Vector : it is a type of physical quantity which will be defined by magnitude and direction both
e) Position : it is defined as the distance with respect to a given reference or origin
f) Acceleration : it is defined by rate of change in velocity.
g) Force = it is push or pull on an object which will create acceleration to the object
F = ma
it is product of mass and acceleration
h) Newton : it is SI unit of force or we measure force in this unit
Answer:
(a)
(b)
Explanation:
Given:
*p = charge on proton =
*e = magnitude of charge on an electron =
*r = distance between the proton and the electron in the Rutherford's atom =
Part (a):
Since two unlike charges attract each other.
According to Coulomb's law:
Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is .
Part (b):
Potential energy between two charges separated by a distance r is given by:
So, the potential energy between the electron and the proton of the Rutherford's atom is given by:
Hence, the electrostatic potential energy of the atom is .