Answer:
Explanation:
Given that
As both charges are negative so there exist force of repulsion in direction as shown in figure.
Angle at which force F12 is acting is
Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction
Answer:
4.763 × 10⁶ N/C
Explanation:
Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.
Resolving E₂ into horizontal and vertical components, we have
E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.
Summing the horizontal components we have
E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²
= -k/r²(q₁ + q₂cos60)
= -k/r²(4 μC + (-6.0 μC)(1/2))
= -k/r²(4 μC - 3.0 μC)
= -k/r²(1 μC)
= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²
= -9 × 10⁵ N/C
Summing the vertical components, we have
E₄ = 0 + (-E₂sin60)
= -E₂sin60
= -kq₂sin60/r²
= -k(-6.0 μC)(0.8660)/(0.10 m)²
= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²
= 46.77 × 10⁵ N/C
The magnitude of the resultant electric field, E is thus
E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴
= 476.28 × 10⁴ N/C
= 4.7628 × 10⁶ N/C
≅ 4.763 × 10⁶ N/C
Answer:
work=f(costheta)
Explanation:
work is done when a force acts on a body and displaces it on the direction of force
Answer:
Initial velocity = 10 m/s
θ = 60°
This is the case of projectile motion
So the horizontal component of velocity 10 m/s = 10 cosθ
u = 10 cosθ
u = 10 cos 60°
u=5 m/s
x= 5 m
So in the horizontal direction
x = u .t
5 = 5 .t
t = 1 sec The vertical component of velocity 10 m/s = 10 sinθ
Vo= 10 sinθ
Vo= 10 sin 60°
Vo = 8.66 m/s
h=3.75 m
So height of robot = 3.75 - 0.75 m
height of robot =3 m
Answer:
No, it's not there.
Explanation:
For a machine to be 100% efficient, it has to be with an output which is equal to its input. But machines have an out put less than an input, hence efficiency below 100%.
