If a 5000-kg is moving at a speed of 43 m/s, what is its momentum?
1 answer:
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Answer:
t = 2.77 s
Explanation:
The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:
Equation of movement of the ball in the X axis
X = v₀x*t Equation (1)
Equation of movement of the ball in the Y axis
Y = y₀+ v₀y*t -½ g*t² Equation (2)
Where
X : horizontal position in meters (m)
Y : vertical position in meters (m)
y₀ : initial vertical position in meters (m)
v₀x : X-initial speed in m/s
v₀y : Y-initial speed in m/s
g: acceleration due to gravity in m/s²
t : time to position (X,Y)
Data
y₀ = 37.5 m
v₀y = 0
g = 9.8 m/s²
Problem development
The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0
We apply the Equation (2):
Y = y₀+ (v₀y)*t - (½) g*t²
0 = 37.5 +(0)*t- (1/2)*(g)*t²
0 = 37.5 - (1/2)*(9.8)*t²
(1/2)*(9.8)*t² = 37.5
t² = (2)(37.5)/(9.8)
t = 2.77 s
Answer:
124.86 V
Explanation:
We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft
1 ft = 0.3048 m
358 ft = 109.12 m
The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m
The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²
Resistivity of wire (ρ) = 0.0171 Ω.mm²/m
The resistance of the wire =
The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V
The voltage at end = 125.2 - 0.34 = 124.86 V