Answer:
The correct Answer to the question is : e) 4.1 m/s^2, 52 degrees north of east
Explanation:
F1= 68 N < 24º = 62.12 i + 27.65 j
F2= 32N < 132º = -21.41 i + 23.78 j
R= F1+F2= 40.71 i +51.43 j = 65.59 N < 51.63 º
By 2nd law of newton:
F= m * a
R= m*a
a= R/m
a= 4.1 m/s² < 52º (52 degrees north of east)
I consideer 0º at the EAST axis.
Answer:
80m<em>/</em><em>s</em>
Explanation:
Final velocity is given by
v=u+at
when a motorcyclist starts from rest, initial velocity (u) =0
therefore
v=0+4*20
v=80m/s
that's the answer
Answer:
As it moves farther and farther away from Q, it's speed will keep increasing.
Explanation:
This is so because the point charges are both positive charges and positive charges repel.
As the charges repel, they do so at an increasing speed.
It is FALSE.
The absolute value of both numbers is 7, therefore they are equal!
Hope this helps!
Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
Let,
m
A
= Mass of ball A
m
B
= Mass of ball B
u
A
= initial velocity of ball A
u
B
= initial velocity of ball B
v
A
= Velocity after the collision of ball A
v
B
= Velocity after the collision of ball B
F
ab
= Force exerted by A on B
F
ba
= Force exerted by B on A
Now,
Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision
= m
A
v
A
−m
A
u
A
Rate of change of momentum A= Change in momentum of A/ time taken
=
t
m
A
v
A
−m
A
u
A
Force exerted by B on A (F
ba
);
F
ba
=
t
m
A
v
A
−m
A
u
A
........ [i]
In the same way,
Rate of change of momentum of B=
t
m
b
v
B
−m
B
u
B
Force exerted by A on B (F
ab
)=
F
ab
=
t
m
B
v
B
−m
B
u
B
.......... [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
F
a
b=−F
b
a [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]
Using [i] and [ii] , we have
t
m
B
v
B
−m
B
u
B
=−
t
m
A
v
A
−m
A
u
A
m
B
v
B
−m
B
u
B
=−m
A
v
A
+m
A
u
A
Finally we get,
m
B
v
B
+m
A
v
A
=m
B
u
B
+m
A
u
A
This is the derivation of conservation of linear momentum.
