Why we don’t say degrees Kelvin. For example Water freezes at 273K .
1 answer:
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Given parameters:
Mass on earth = 50kg
Unknown:
Mass on planet Xenon = ?
Weight on planet Xenon = ?
Mass is the amount of matter contained in a particular substance.
Weight is the force on a body and it is derived from the product of mass and acceleration due to gravity.
Weight = mass x acceleration due to gravity
Planet Xenon has half the gravitational force of Earth.
This translated gives = 4.9m/s²
Now, mass is always the same every where if the amount of matter in a substance does not change.
In this problem, mass = 50kg on planet xenon.
Weight = mass x acceleration due to gravity = 50 x 4.9 = 245N
The weight on Xenon is 245N and the mass is 50kg
Answer:
1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s
Explanation:
1) First Scenario.
After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.
Before starting, let's reduce the magnitudes to the SI system
v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s
y = 2650 m
Let's start by looking for the time it takes for the load to reach the ground.
y = y₀ + v_{oy} t - ½ g t²
in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero
0 = y₀ + 0 - ½ g t2
t =
t = √(2 2650/9.8)
t = 23.26 s
this is the horizontal scrolling time
x = v₀ t
x = 366.67 23.26
x = 8527 m
the speed at the point of arrival is
v_y = v_{oy} - g t = 0 - gt
v_y = - 9.8 23.26
v_y = -227.95 m / s
Module and angle form
v =
v = √(366.67² + 227.95²)
v = 431.75 m / s
θ = tan⁻¹ (v_y / vₓ)
θ = tan⁻¹ (227.95 / 366.67)
θ = - 31.97º
measured clockwise from x axis
We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.
2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º
we look for the components of speed
cos θ = v₀ₓ / v₀
sin θ = v_{oy} / v₀
v₀ₓ = v₀ cos θ
v_{oy} = v₀ sin θ
we look for the time for the arrival point that has coordinates x = 0, y = 0
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + vo sin θ t - ½ g t²
0 = -50 + vo sin 50 t - ½ 9.8 t²
x = x₀ + v₀ₓ t
0 = x₀ + vo cos θ t
0 = -400 + vo cos 50 t
podemos ver que tenemos un sistema de dos ecuación con dos incógnitas
50 = 0,766 vo t – 4,9 t²
400 = 0,643 vo t
resolved
50 = 0,766 ( ) t – 4,9 t²
50 = 476,52 t – 4,9 t²
t² – 97,25 t + 10,2 = 0
we solve the quadratic equation
t = [97.25 ± ] / 2
t = 97.25 ±97.04] 2
t₁ = 97.145 s
t₂ = 0.1 s≈0
the correct time is t1 the other time is the time to the launch point,
t = 97.145 s
let's find the initial velocity
x = x₀ + v₀ cos 50 t
0 = -400 + v₀ cos 50 97.145
v₀ = 400 / 62.44
v₀ = 6.4 m / s
