What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the p
1 answer:
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Answer:
vo = 227 m/s
Explanation:
Let's analyze the problem, we have two parts one during the crash for which we will use the moment and another when it slides where we will use the energy. As we give the data of the latter let's start here
Before starting let's reduce the unit to the SI system
m = 6.00 g (1kg / 1000g) = 6.00 10⁻³ kg
M = 1.12 kg
We write the energy at two points, one initial right after the crash and another when the body has stopped
Just after shock
The bodies are united, so the mass is the sum of the mass of the bullet and the block
Em₀ = K = ½ (m + M) v²
When the body has stopped
= 0
When in the system there is friction force the variation the mechanical energy is equal to the work of the friction force. Notice that the force of friction opposes the movement so that their work is negative
W fr = ΔEm = -Em₀
-fr d = 0 - ½ (m + M) v²
To find the force of friction let's use Newton's second law
Axis y
N-W = 0
N = W = (m + M) g
The equation for the force of friction is
fr = μ N
fr = μ (m + M) g
Let's replace in the work and energy equation
-μ (m + M) g d = 0 - ½ (m + M) v²
From here we can find the system speed. Let's calculate
v³ = 2 μ g d
v = √ (2μ gd)
v = √ (2 0.250 9.8 0.300)
v = 1.21 m / s
Now let's solve the crash, let's look for the moment before and after it
Before the crash
po = m vo
After the crash
= (m + M) v
The system is formed by the bullet and the block, therefore, the forces during the impact are internal and the amount of movement is conserved
po =
m vo = (m + M) v
vo = v (m + M) / m
vo = 1.21 (0.00600 + 1.12) /0.00600
vo = 227 m / s
Answer:
Mirages happen when the ground is very hot and the air is cool.
Explanation:
They happen when light passes through two layers of air with different temperatures. The desert sun heats the sand, which in turn heats the air just above it. The hot air bends light rays and reflects the sky.
When you see it from a distance, the different air masses colliding with each other act as a mirror.
Answer:
a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m
Explanation:
a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.
So y - y₀ = ut - 1/2gt²
y - y₀ = (v₀sinθ)t - 1/2gt²
substituting the values of the variables into the equation, we have
0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²
- 60 = 50t - 4.9t²
So, 4.9t² - 50t - 60 = 0
Using the quadratic formula to find t,
Since t cannot be negative, t = 11.29 s
b. We first need to find the impact vertical velocity component. Using
v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,
v = v₀sinθ - gt
= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s
= 50.01 m/s - 110.64 m/s
= -60.63 m/s
Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.
The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)
= √((-60.63 m/s)² + (72.77 m/s)²)
= √((3676 m²/s² + 5295.48 m²/s²)
= √(8971.48 m²/s²
= 94.72 m/s
The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°
So the impact velocity is 94.72 m/s at -39.8°
c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m