7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
Answer:
Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force
Explanation:
at least part of which is applied in the direction of the displacement. ... To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd.
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
<h2>When Transferring Flammable Liquids - Option A</h2>
To create a pathway for the liquid to flow easily is a correct reason to physically connect objects with a bond wire when transferring flammable liquids. Physically connect the two conductive objects together with a bond wire to eliminate a difference in static charge potential among them. The bond wire is attached between the containers in the process of flammable filling liquid actions except for a metallic path is found.
