a)., b)., and c). are completely false.
There isn't a grain of truth among them.
In Physics, the technical definition of 'Work' is (force) times (distance).
Answer:
1.86 m
Explanation:
First, find the time it takes to travel the horizontal distance. Given:
Δx = 52 m
v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s
a = 0 m/s²
Find: t
Δx = v₀ t + ½ at²
52 m = (22.2 m/s) t + ½ (0 m/s²) t²
t = 2.35 s
Next, find the vertical displacement. Given:
v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s
a = -9.8 m/s²
t = 2.35 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²
Δy = 4.91 m
The distance between the ball and the crossbar is:
4.91 m − 3.05 m = 1.86 m
Answer:
one-third of its weight on Earth's surface
Explanation:
Weight of an object is = W = m*g
Gravity on Earth = g₁ = 9.8 m/s
Gravity on Mars = g₂ =
g₁
Weight of probe on earth = w₁ = m * g₁
Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )
As g₂ = g₁/3 --------- ( 2 )
Put equation (2) in equation (1)
so
Weight of probe on Mars = w₂ = m * g₁ /3
Weight of probe on Mars =
m * g₁ =
w₁
⇒Weight of probe on Mars =
Weight of probe on earth
Answer:
v_average = 15 m / s
Explanation:
The average speed can be found in two ways,
* taking the distance traveled and divide it by the time spent
* taking the velocities in each time interval and then finding the weighted average by the time fraction
v_average = 1 / t_total ∑
vi ti
Let's apply this last equation
Total time is
t = t₁ + t₂
t = 10 + 10 = 20 min
v_average = 10/20 10 + 10/20 20
v_average = 10/2 + 20/2
v_average = 15 m / s