On the EM spectrum, visible light is found between
2 answers:
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Answer: -3.49 m/s (to the south)
Explanation:
This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum must be equal to the final momentum
, and taking into account this is aninelastic collision:
Before the collision:
(1)
After the collision:
(2)
Where:
is the mass of the car
is the velocity of the car, directed to the north
is the mass of the truck
is the velocity of the truck, directed to the south
is the final velocity of both the car and the truck
(3)
(4)
Isolating :
(5)
(6)
Finally:
The negative sign indicates the direction of the velocity is to the south
Answer:
h = 10.4 m
R = 22.48 m
v= 16,2 m/s , α = 61.7°, below the horizontal
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)
Explanation:
The ball describes a parabolic path, and the equations of the movement are:
Equation of the uniform rectilinear motion (horizontal ) :
x = vx*t :
Equations of the uniformly accelerated rectilinear motion of upward (vertical ).
y = (v₀y)*t - (1/2)*g*t² Equation (2)
vfy² = v₀y² -2gy Equation (3)
vfy = v₀y -gt Equation (4)
Where:
x: horizontal position in meters (m)
t : time (s)
vx: horizontal velocity in m
y: vertical position in meters (m)
v₀y: initial vertical velocity in m/s
vfy: final vertical velocity in m/s
g: acceleration due to gravity in m/s²
Known data
y= 8.8 m
v = ( (7.7)i + (5.7)j ) m/s : vx= 7.7 m/s , vy= 5.7 m/s
g = 9.8 m/s²
Calculation of the initial vertical velocity ( v₀y)
We apply Equation (3) with the known data
(vfy)² = (v₀y)² -2*g*y
(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)
(5.7)²+ 172.48 = (v₀y)²
v₀y = 14.3 m/s
Calculation of the maximum height the ball rise (h)
In the maximum height vfy=0
We apply the Equation (3) :
(vfy)² = (v₀y)² -2*g*y
0 = (14.3)² - 2*98*h
h = (14.3)² / 19.6
h = 10.4 m
Calculation of the time it takes for the ball to the maximum height
We apply the Equation (4) :
vfy = v₀y -gt
0 = v₀y -gt
gt = v₀y
t = v₀y/g
t = 14.3/9.8
t= 1.46 s
Flight time = 2t = 2.92 s
Total horizontal distance traveled by the ball (R)
We replace data in the equation (1)
x =vx*t vx= 7.7 m/s , t =2.92 s (Flight time)
R = (7.7)* (2.92) = 22.48 m
Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground
vx = 7.7 m/s
vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s
v= 16,2 m/s
α = -61.7°
α = 61.7°, below the horizontal
i- j components of the v
v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)