Answer:
Energy expenditure in K cals/min = 10 K cals /min (approximately)
Explanation:
As we know
Energy expenditure in Kcal/min= METs x 3.5 x Body weight (kg) / 200
Given is METs=7.6
Weight of Jazz= 172lb=78.02kg
putting the values in formula,
Energy expenditure in K cals/min= 7.6 x 3.5 x 78.02 / 200
=10.38 K cals /min
=10 K cals /min (approximately)
Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min
Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m
If the net force is 4 N, and Frankie is pulling the rope with 7 N, Carol must be pulling the rope with 11 N (I think that Carol is going to win the tug-of-war...).