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


Answer 1


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 = (m)/((I_o)/(r)  +mr) \ r v_o

let's calculate

       v = (0.020)/((1.4)/(0.6  ) + 0.020 \ 0.6  ) \ 0.6 \ 4

       v = (0.020)/(2.345) \ 2.4

       v = 0.02 m / s


To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s


         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

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A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.

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?



D=1.54489 m


Given data

S=6.10 mm= 0.0061 m

To find

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

S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n  t=\sqrt{1.245*10^(-3) }\n t=0.035s

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

v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s

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

Where ;

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

such as:

  • 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

they are much active.

Read more about Ojibwe people here

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


A.) Sunny


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)




solution below

Final answer:

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.

Learn more about Quantum Energy here:


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

v^2-u^2=2as\n\Rightarrow -u^2=2as-v^2\n\Rightarrow u=√(v^2-2as)\n\Rightarrow u=√(0^2-2* -9.81* 1)\n\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\n\Rightarrow t=(v-u)/(a)\n\Rightarrow t=(0-4.45)/(-9.81)\n\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+(1)/(2)at^2\n\Rightarrow 1=0t+(1)/(2)* 9.81* t^2\n\Rightarrow t=\sqrt{(1* 2)/(9.81)}\n\Rightarrow t=0.45\ s

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



56.52 feet³ to the nearest hundredth


the volume of a cone is given as

V =

(1)/(3) \pi r^(2) h

the radius is 3 feet

height is 6 feet

substituting this given values in the formular

we have that, V = (1)/(3) x 3.14 x 3^(2) x 6

dividing , we have the volume (V)

V= 3.14 x 3 x 6

= 3.14 x 18

= 56.52 feet³ to the nearest hundredth