Answer:
B. 1500 kg*m/s
Explanation:
Momentum p = m* v
In any type of collision, the total momentum is preserved!
The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.
p1 + p2 =
m1*v1 + m2*v2
m1 = me = 300 kg
v1 = 3 m/s
v2 = 2 m/s
Substitute the given numbers:
300*3 + 300+2
900 + 600
1500 kg*m/s, which is answer B.
It depends on what they are
Answer:
b) Nothing will happen, the sea saw will still be balanced.
Explanation:
b) Nothing will happen, the sea saw will still be balanced.
Reason:-
When two kids are balanced, the sum of torques on the seesaw will be zero.
if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.
Therefore the seesaw is balanced
Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
The most common form of angina is stable angina.
C. stable
