Answer:
W = -1844.513 J
Explanation:
GIVEN DATA:
mass of spider man is m 74 kg
vertical displacement if spider is 11 m
final displacement = 11 cos 60.6 = - 6.753 m
change in displacement is = -6.753 - (-11) = 4.25 m
gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N
work done by gravity is
where 180 is the angle between spiderman weight and displacement
W = -1844.513 J
Answer:
Market segmentation helps you send the right message, every time, by efficiently targeting specific groups of consumers.
Explanation:
Unscrambling
1. resting heart rate
2. overload
3. workout
4. specificity
5. cool-down
6. progression
7. warm-up
8. the last one can only be instance, but there was a typo on the paper.
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
Thermal equilibrium is a state in which all parts of a system are at the same temperature