AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.
Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law
F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law
F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
