consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
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A. Internal. Most cars use that type of set up because it's more efficient, you can find more about it on this website, https://auto.howstuffworks.com/did-cars-ever-have-external-combustion-engines.htm
:)
~ Ria
Answer:
Explanation:
a. The equation of Lorentz transformations is given by:
x = γ(x' + ut')
x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.
x' = 0
t' = 5.00 s
u =0.800 c,
c is the speed of light = 3×10⁸ m/s
Then,
γ = 1 / √ (1 - (u/c)²)
γ = 1 / √ (1 - (0.8c/c)²)
γ = 1 / √ (1 - (0.8)²)
γ = 1 / √ (1 - 0.64)
γ = 1 / √0.36
γ = 1 / 0.6
γ = 1.67
Therefore, x = γ(x' + ut')
x = 1.67(0 + 0.8c×5)
x = 1.67 × (0+4c)
x = 1.67 × 4c
x = 1.67 × 4 × 3×10⁸
x = 2.004 × 10^9 m
x ≈ 2 × 10^9 m
Now, to find t we apply the same analysis:
but as x'=0 we just have:
t = γ(t' + ux'/c²)
t = γ•t'
t = 1.67 × 5
t = 8.35 seconds
b. Mavis reads 5 s on her watch which is the proper time.
Stanley measured the events at a time interval longer than ∆to by γ,
such that
∆t = γ ∆to = (5/3)(5) = 25/3 = 8.3 sec which is the same as part (b)
c. According to Stanley,
dist = u ∆t = 0.8c (8.3) = 2 x 10^9 m
which is the same as in part (a)
