Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1
Answer:
A.) 27000 kgm/s
18000 kgm/s
B.) Va = 22 m/s
C.) 19800 kgm/s
25200 kgm/s
Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g
M = 9×10^5/1000 = 900 kg
A.) Initial momentum of A
Mu = 900 × 30 = 27000 kgm/s
Initial momentum of B
Mu = 900 × 20 = 18000 kgm/s
B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.
Momentum before impact = momentum after impact
Given that Vb = 28 m/s
27000 + 18000 = 900Va + 900 × 28
45000 = 900Va + 25200
900Va = 45000 - 25200
900Va = 19800
Va = 19800/900
Va = 22 m/s
C.) Momentum of A after impact
MV = 900 × 22 = 19800 kgm/s
Momentum of B after impact
MV = 900 × 28 = 25200 kgm/s
Answer:
Explanation:
The horizontal component of force applied = 7.97 cos 26.2 = 7.15 N.
Vertical component in upward direction = 7.97 sin 26.2 = 3.52 N.
Since the body is moving with uniform velocity, friction force will equalize the external horizontal component = 7.15 N.
So frictional force = 7.15 N.
a) Work done by rope force per second = force in horizontal direction x displacement per second = 7.15 x 4.33 = 30.95 J
b) Increase in thermal energy per second will be due to negative work done by frictional force = work done by external force = 30.95 J.
c) Normal force acting downwards = weight - vertical component of external force = 4.36 x 9.8 - 3.52 = 39.21 N
coefficient of friction = friction force / normal force = 7.15 / 39.21 = 0.18
Answer:
answered it.................
