Answer:
50 N
Explanation:
Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:
The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.
The horizontal component of the forces:
F₁ + F₂ = -40N + F₂ = 0
F₂ = 40N
The vertical component of the forces:
F₁ + F₂ - mg = 0 + F₂ - mg = 0
F₂ = mg
If I assume the gravitational constant g = 10 m/s²:
F₂ = (3 kg) * (10 m/s²) = 30N
Adding the horizontal and vertical components of the force F₂:
F₂ = √((40N)² + (30N)²) = 50N
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
Answer:
1.56 J
Explanation:
The potential energy only depends on the vertical height from the ground level.
We consider the ground level to have zero P.E.
So when it is 2 m above the ground level,
P.E. = mgh
= 0.078×10×2
= 1.56 J
2.71 m/s fast Hans is moving after the collision.
<u>Explanation</u>:
Given that,
Mass of Jeremy is 120 kg (
)
Speed of Jeremy is 3 m/s (
)
Speed of Jeremy after collision is (
) -2.5 m/s
Mass of Hans is 140 kg (
)
Speed of Hans is -2 m/s (
)
Speed of Hans after collision is (
)
Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is
= 
Substitute the given values,
= 120 × 3 + 140 × (-2)
= 360 + (-280)
= 80 kg m/s
Linear momentum after the collision of Jeremy and Hans is
= 
= 120 × (-2.5) + 140 ×
= -300 + 140 ×
We know that conservation of liner momentum,
Linear momentum before the collision = Linear momentum after the collision
80 = -300 + 140 ×
80 + 300 = 140 ×
380 = 140 ×
380/140=
= 2.71 m/s
2.71 m/s fast Hans is moving after the collision.
Answer:
80 angle of incidence=angle of reflection
Explanation:
