Answer:
Explanation:
<u>Average Acceleration
</u>
Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.
Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that.
</em>
When 
, the graph shows a value of 
When 
, the object is at rest, 
We compute the average acceleration as
Preserved fossil<span> (like a fossil in amber, ice or tar.</span>
Answer:
The maximum height reached by the ball is 16.35 m.
Explanation:
Given;
initial velocity of the ball, u = 17.9 m/s
the final velocity of the ball at the maximum height, v = 0
The maximum height reached by the ball is given by;
v² = u² + 2gh
During upward motion, gravity is negative
v² = u² + 2(-g)h
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u² / 2g
h = (17.9)² / (2 x 9.8)
h = 16.35 m
Ttherefore, the maximum height reached by the ball is 16.35 m.
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)
Answer:
1. You push on the ball and the ball pushes on your hand
.
2. The ball hits the ground and the ground pushes back on the ball
.
3. You walk on the ground with your feet and the ground pushes back on you.
Explanation:
