For a human jumper to reach a height of 110 cm, the person will need to leave the ground at a speed of 4.65 m/s.
We can calculate the initial speed to reach 110 cm of height with the following equation:
Where:
: is the final speed = 0 (at the maximum height of 110 cm)
: is the initial speed =?
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 110 cm = 1.10 m
Hence, the <u>initial velocity</u> is:
Therefore, the initial speed that the person must have to reach 110 cm is 4.65 m/s.
You can see another example here: brainly.com/question/13359681?referrer=searchResults
I hope it helps you!
Answer:
because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.
Answer: Remember speed is distance divided by time, so if he travels 1000 m in 7.045 s, his speed is
(1000 m)/(7.045 s) = 141.9 m/s.
Note there are 1609 metres in a mile, or 1 mi = 1609 m, so m = 1/1609 mi, or
141.9/1609 mi/s = 0.08822 mi/s. Now, note that 1 h = 3600 s, so the speed is
0.08822*3600 mi/h = 317.6 mi/h.
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)
Explanation:
We define force as the product of mass and acceleration.
F = ma
It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.
Given Data:
Width of the pool = w = 50 ft
length of the pool = l= 100 ft
Depth of the shallow end = h(s) = 4 ft
Depth of the deep end = h(d) = 10 ft.
weight density = ρg = 62.5 lb/ft
Solution:
a) Force on a shallow end:
b) Force on deep end:
c) Force on one of the sides:
As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.
1) Force on the Rectangular part:
2) Force on the triangular part:
here
h = h(d) - h(s)
h = 10-4
h = 6ft
now add both of these forces,
F = 25000lb + 150000lb
F = 175000lb
d) Force on the bottom:
