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:
0.073 N-m
Explanation:
i = 12 A, l = 0.8 m, B = 0.12 T
The circumference of the loop is 0.8 m.
Let r be the radius of the loop.
2 x 3.14 x r = 0.8
r = 0.127 m
Maximum Torque = i x A x B
Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m
Answer:
Once you know the differences between meiosis 1 and 2, you will remember it easier.
Explanation:
Meiosis 1 starts with 1 diploid cell Meiosis 2 starts with 2 haploid cells,
each with a homologous pair
Meiosis 1 results in 2 daughter cells Meiosis 2 results in 4
17 Sin (45) in the vertical and 17 Cos
(45)
<span>Just add the two kinetic energies;
E = (1/2)mv^2 + (1/2)mv^2</span>
