a liquid with a specific heat of 1900J/Kg° C has 4750 joules of heat energy is added to it. if the temperature increases from 20
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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)