Answer:
For a given spring the extension is directly proportional to the force applied For example if the force is doubled, the extension doubles When an elastic object is stretched beyond its limit of proportionality the object does not return to its original length when the force is removed
Explanation:
Answer:
W₃ = 3310.49 J
, W3 = 3310.49 J
Explanation:
We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections
We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics
v2 = v₀² + 2 a₁ y
as they rest part of the rest the ricial speed is zero
v² = 2 a₁ y
a₁ = v² / 2y
a₁ = 2.3² / (2 5.90)
a₁ = 0.448 m / s²
with this acceleration we can calculate the applied force, using Newton's second law
F -W = m a₁
F = m a₁ + m g
F = m (a₁ + g)
F = 69 (0.448 + 9.8)
F = 707.1 N
Work is defined by
W₁ = F.y = F and cos tea
As the force lifts the man, this and the displacement are parallel, therefore the angle is zero
W₁ = 707.1 5.9
W₁ = 4171.89 J W3 = 3310.49 J
Let's calculate for the second part
the speed is constant, therefore they relate it to zero
F - W = 0
F = W
F = m g
F = 60 9.8
F = 588 A
the job is
W² = 588 5.9
W2 = 3469.2 J
finally the third part
in this case the initial speed is 2.3 m / s and the final speed is zero
v² = v₀² + 2 a₂ y
0 = vo2₀² + 2 a₂ y
a₂ = -v₀² / 2 y
a₂ = - 2.3²/2 5.9
a2 = - 0.448 m / s²
we calculate the force
F - W = m a₂
F = m (g + a₂)
F = 60 (9.8 - 0.448)
F = 561.1 N
we calculate the work
W3 = F and
W3 = 561.1 5.9
W3 = 3310.49 J
total work
W_total = W1 + W2 + W3
W_total = 4171.89 +3469.2 + 3310.49
w_total = 10951.58 J
Answer:
u/2 √(1 + 3 cos² θ)
Explanation:
The object is thrown at an angle θ, so the velocity has two components, vertical and horizontal.
Initially, the vertical component is u sin θ and the horizontal component is u cos θ.
At the maximum height, the vertical component is 0 and the horizontal component is u cos θ.
The mean vertical velocity is:
(u sin θ + 0) / 2 = u/2 sin θ
The mean horizontal velocity is:
(u cos θ + u cos θ) / 2 = u cos θ
The net mean velocity can be found with Pythagorean theorem:
v² = (u/2 sin θ)² + (u cos θ)²
v² = u²/4 sin² θ + u² cos² θ
v² = u²/4 (1 − cos² θ) + u² cos² θ
v² = u²/4 (1 − cos² θ) + u²/4 (4 cos² θ)
v² = u²/4 (1 − cos² θ + 4 cos² θ)
v² = u²/4 (1 + 3 cos² θ)
v = u/2 √(1 + 3 cos² θ)
86,400 is the amount of seconds in a day. :)
It’s a speed of 15 m/s so it would be their velocities after collision
