First you need to calculate the velocity of the stone when it reaches the ground level. This is easy to find from energy conservation, since the potential energy it had at the top of the tower has been totally converted into kinetic energy.
We don't know the mass of the stone, but it cancels from both equations. This gives
so v=44.27 m/s.
Now, the time it took the stone to fall from the top of the tower to the ground is calculated easily from
.
The initial velocity is 0, the initial height is 100 meters and the final height is 0 since we are taking the ground floor as height 0.
This gives
So the time it took the stone to fall from the ground level to the bottom of the well is 5-4.52=0.48 s.
We can now use
where v is the velocity we calculated before v=44.27 m/s, time is t=0.48 s and xf is the depth of the well.
So the solution is 22.5 m
Answer:
ΔX = 0.0483 m
Explanation:
Let's analyze the problem, the car oscillates in the direction y and advances with constant speed in the direction x
The car can be described with a spring mass system that is represented by the expression
y = A cos (wt + φ)
The speed can be found by derivatives
= dy / dt
= - A w sin (wt + φ
So that the amplitude is maximum without (wt + fi) = + -1
= A w
X axis
Let's reduce to the SI system
vₓ = 15 km / h (1000 m / 1 km) (1h / 3600s) = 4.17 m / s
As the car speed is constant
vₓ = d / t
t = d / v
ₓ
t = 4 / 4.17
t = 0.96 s
This is the time between running two maximums, which is equivalent to a full period
w = 2π f = 2π / T
w = 2π / 0.96
w = 6.545 rad / s
We have the angular velocity we can find the spring constant
w² = k / m
m = 1200 + 4 80
m = 1520 m
k = w² m
k = 6.545² 1520
k = 65112 N / m
Let's use Newton's second law
F - W = 0
F = W
k x = W
x = mg / k
Case 1 when loaded with people
x₁ = 1520 9.8 / 65112
x₁ = 0.22878 m
Case 2 when empty
x₂ = 1200 9.8 / 65112
x₂ = 0.18061 m
The height variation is
ΔX = x₁ -x₂
ΔX = 0.22878 - 0.18061
ΔX = 0.0483 m
Answer:
Explanation:
a) See the attached file for diagram . The two triangles are ABC and DEC .
b )
Angle of incidence = i
Tan i = ( 309 - 227 ) / 105
= 82 / 105 = .78
i = 38°
c )
Sin i / sinr = 1 / μ where μ is refractive index of water .
Sin i / sinr = 1 / μ = 1 / 1.33
sin 38 / sinr = .7518
.6156 / sinr = .7518
sinr = .6156 / .7518
.819
r = 55°
d )
angle the refracted beam makes with the horizontal = 90 - 55 = 35°
e )
h / 227 = Tan 35
h = 227 x Tan 35
= 159 m
Height required = 159 m .
This is how you calculate his speed, for the 100metres. That is, the average speed for the entire distance, including the start from intertia until the race end where the athlete is decelerating. Multiply the time by 10, and divide 3600 by that figure. i.e. 10 by10 equals 100. Divide 3600 by 100 and you get 36 which is the speed figure you requested.
A female athlete ran 200 metres a few years ago and in the middle section of the race covered 100 metres in that race in a time under 10 seconds. That is, she ran a sub 10second 100metres from a flying start. This great athlete from Slovenia, also ran a 100 metre race (from a stationary start) in a time under 11 seconds, when she was 48 years of age. Her name is Merlene Ottey.
Answer:
C.
Explanation:
Of the following activity best demonstrates the creativity of Rutherford: is making observations about the gold foil experiment.
Rutherford's Gold Foil Test confirmed the existence of a thin, enormous atom core that would later be recognized as an atom's nucleus. To examine the effect of alpha particles on material, Ernest Rutherford, Hans Geiger and Ernest Marsden conducted their Gold Foil Experiment.