An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?a. 1min b. 2min c. 3min d. 4min e. 5min f. 6min
T = 188.5 s, correct is C
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?
S=6.10 mm= 0.0061 m
Depth of lake
To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation
So ball takes 0.035sec to hit the water
As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as
Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake
So the depth of lake given as:
Answer: d = 1.54m
The depth of the lake is 1.54m
The final velocity of the ball just before it hit the water can be derived using the equation below;
v^2 = u^2 + 2as ......1
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance travelled.
Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:
v^2 = 2gs
v = √2gs ......2
g = 9.8m/s^2
s = 6.10mm = 0.0061m
substituting into equation 2
v = √(2 × 9.8× 0.0061)
v = 0.346m/s
The time taken for the ball to hit water from the time of release can be given as:
d = ut + 0.5gt^2
Since u = 0
d = 0.5gt^2
Making t the subject of formula.
t = √(2d/g)
t = √( 2×0.0061/9.8)
t = 0.035s
The time taken for the ball to reach the bottom of the lake from the when it hits water is:
t2 = 4.5s - 0.035s = 4.465s
And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;
depth d = velocity × time = 0.346m/s × 4.465s
d = 1.54m
The depth of the lake is 1.54m
Why must the Ojibwe people pay close attention to the seasons? a.) they must be ready to move to a new place where they can hunt
b.) they only fish during the warmest times of the day
c. they must know the right time of year for Gathering certain foods
d.) they still catch walleye with the steering method
The Ojibwe people pay close attention to the seasons in order to know right
time of year for gathering certain foods.
The Ojibwe mostly hunt for fishes through the use of various techniques
Fishing at night
Use of flashlight
Why do they hunt for Fishes at night?
They hunt for fishes at night because they are usually docile during that time
which enables them to catch them easily as against during the day when
Answer:C. They must know the right time of year for gathering certain foods
I got it correct
A local meteorologist reports the day’s weather. "Currently sunny outside, 34°F. Skies will become overcast later this afternoon, as temperatures drop to 25°F, with windy conditions out of the north at 10–15 miles per hour. Radar indicates 2–3 inches of snow expected to fall later tonight.” Which information is qualitative? These are non-numerical, descriptive data. These are numerical data that have been measured. “sunny” “25°F” “2–3 inches of snow” “10–15 miles per hour”
took the test
The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)
The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.
To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.
One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.
The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.
Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 100 cm . a) What is their initial "launch" speed off the ground?b)How long are they in the air?
a) 4.45 m/s
b) 0.9 seconds
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
a) The vertical speed when the player leaves the ground is 4.45 m/s
Time taken to reach the maximum height is 0.45 seconds
Time taken to reach the ground from the maximum height is 0.45 seconds
b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds
What is the volume of a cone with a radius of 3 feet and a height of 6 feet use 3.14 for pie round your answer to the nearest hundredth