5. A big wheel has a diameter of 5 m and a mass of 1500 kg when fully laden with people. a) Work out the moment of inertia of the big wheel. (Hint: which shape from the ones given on p114 would be most suitable? b) When the wheel is rotating at full speed, a person has a linear velocity of 3 m/s. What is the angular velocity of this person? c) What is the rotational kinetic energy at this speed? d) A motor takes 10 seconds to accelerate the wheel from rest to a linear velocity on the circumference of 3 m/s. What is the power of the motor?​

Explanation:

mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

We have that  the ratio of his walking pace  to the walking pace of a person of average height is

[tex]\frac{V_2}{V_1}=1.10[/tex]

given the assumption and the calculation given below

From the question we are told that:

Consider a person 18\% shorter than average

Let average height of a person be [tex]10m[/tex]

Therefore

The height of an [tex]18\%[/tex] shorter man is mathematically given as

H=10*0.18

H=8.2m

Generally, the equation for velocity is mathematically given by

[tex]v=\frac{1}{2\pi} \sqrt{{g}{l}}[/tex]

Where we have the  Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same

Therefore

[tex]\frac{V_1}{V_2}=\frac{l_1}{l_2}[/tex]

[tex]\frac{V_1}{V_2}={82}{100}[/tex]

[tex]\frac{V_2}{V_1}=1.10[/tex]

In conclusion

The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is

[tex]\frac{V_2}{V_1}=1.10[/tex]

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Answer:

I = 15,645. kg*m/s or 15,645 N*s

Explanation:

I = m(^v)

I = 1050kg((26.2m/s-11.3m/s)

I = 15,645. kg*m/s

Answer:

a) Fnet = mg - Fb - Fr

b) 8.67 secs

Explanation:

mass of object = 80 kg

Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696

Proportionality constant = 10 N-sec/m

a) Calculate  equation of motion of the object

Force of resistance on object  due to water = Fr ∝ V

                                                                         = Fr = Kv = 10 V

Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object  while mg ( weight ) acts in the negative y - direction on the object.

Fnet = mg - Fb - Fr

∴ Equation of motion of the object ( Ma = mg - Fb - Fr )

b) Calculate how long before velocity of the object hits 40 m/s

Ma = mg - Fb - Fr

a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V

V = u + at ---- ( 1 )

u = 0

V = 40 m/s

a = 9.6138 - 0.125 V

back to equation 1

40 = 0 + ( 9.6138 - 0.125 (40) ) t

40 = 4.6138 t

∴ t = 40 / 4.6138 = 8.67 secs

Answer:

(A) Gravity is you're answer.

Explanation:

When an object or human is falling at an increased rate, The force of gravity is taking place.

Answer:

The torque decreases because as the hanging mass goes down, the moment arm about the pivot point decreases. Since the torque is directly proportional to the length of the moment arm, torque decreases.

The torque decreases because as the hanging mass goes down, the moment arm about the pivot point decreases. Since the torque is directly proportional to the length of the moment arm, torque decreases.

Atomic number of an element  is defined as total number of protons present in the nucleus,  neutrons carry no net electrical charge, so it is the charge number of the nucleus.

atomic mass of an element can be defined as the atomic weight is measured total mass of an element’s atom, the total number of neutrons and protons in the nucleus of an atom.

Both Atomic mass and an atomic number of elements are closely related if  atomic number is high, then the atomic mass is also said to be high.

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a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency

Answer:

Explanation:

The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:

initial velocity is 20 m/s

Here's what we know that we are NOT told:

a = -9.8 m/s/s and

final velocity is 0 at an object's max height in parabolic motion.

We will use the equation:

[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:

0 = 20 + (-9.8)t and

-20 = -9.8t so

t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.

The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.

Acceleration:

We can use the fallowing kinematic equation to get the acceleration at 80 s.

[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]            

Where:

The, we have:

[tex]\vec{a}=\frac{60-20}{80}[/tex]

[tex]\vec{a}=0.5\: m/s^{2}[/tex]

Position:

Knowing the acceleration we can find the position using the falling equation.

[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]

[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]

[tex]\vec{x}=3200 m[/tex]

Average velocity:

The definition of the average velocity is:

[tex]\vec{v}=\frac{\Delta x}{t}[/tex]

[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]

[tex]\vec{v}=\frac{3200-0}{80}[/tex]

[tex]\vec{v}=40\: m/s[/tex]

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Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

[tex]L = \frac{N^2 \mu_0 A}{l}[/tex]

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

[tex]A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2[/tex]

[tex]L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH[/tex]

(b) The energy stored in the inductor when 21 A current ;

[tex]E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J[/tex]

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

[tex]emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s[/tex]

Answer:

375 is the answer.

Explanation:

Speed : Distance / Time taken

S: m/ s

s: 1500/4

375 m / s answer

Answer:

375m per minute

Explanation:

if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame

Answer:

the change in the kinetic energy of the system is -42.47 J

Explanation:

Given;

mass A, Ma = 2 kg

initial velocity of mass A, Ua = 15 m/s

Mass M, Mm = 4 kg

initial velocity of mass M, Um = 7 m/s

Let the common velocity of the two masses after collision = V

Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;

[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]

The initial kinetic of the two masses;

[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]

The final kinetic energy of the two masses;

[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]

The change in kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]

Therefore, the change in the kinetic energy of the system is -42.47 J

Answer:

[tex]F=133N[/tex]

Explanation:

From the question we are told that:

Length [tex]l=3.0m[/tex]

Mass [tex]m=24kg[/tex]

Distance from Tip [tex]d=35cm[/tex]

Generally, the equation for Torque Balance is mathematically given by

[tex]mg(l/2)=F(l-d)[/tex]

[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]

Therefore

[tex]F=133N[/tex]

Answer:

[tex]u=34cm[/tex]

Explanation:

From the question we are told that:

Far point is [tex]V=34 cm[/tex]

Near point is [tex]u=17 cm[/tex]

Therefore

Focal Length

[tex]f=-34cm[/tex]

Generally the equation for the Lens is mathematically given by

[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]

[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]

[tex]u=34cm[/tex]

Answer:

rock go down

Explanation:

what comes up must come down.

Explanation:

Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

Answer:

c) kinetic energy

Explanation:

Answer: C)  kinetic energy

Explanation:

Answer:

Explanation:

F = ma

F = (10)(3)

F = 30 N

Answer:

[tex]\boxed {\boxed {\sf 30 \ Newtons}}[/tex]

Explanation:

We are asked to find the applied net force. According to Newton's Second of Law, force is the product of mass and acceleration.

[tex]F= m \times a[/tex]

The object has a mass of 10 kilograms and it is accelerating at 3 meters per second squared.

Substitute the known values into the formula.

[tex]F= 10 \ kg \times 3 \ m/s ^2[/tex]

Multiply.

[tex]F= 30 \ kg \times m/s^2[/tex]

1 kilogram meter per second squared is equal to 1 Newton, so our answer of 30 kg × m/s² is equal to 30 N.

[tex]F= 30 \ N[/tex]

The applied net force is 30 Newtons.

One of the two requirements of electric current is there must be an electric potential between two bodies

For electric current to flow, there must be an electric potential between two bodies.

This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.

The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.

This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.

So, one of the two requirements of electric current is there must be an electric potential between two bodies.

So, the answer is A

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Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

Answer:

The initial velocity is 1.27 m/s.

Explanation:

distance, s = 1.8 m

acceleration, a = - 0.45 m/s^2

final velocity, v = 0

let the initial velocity is u.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]

We have that the Initial velocity  is mathematically given as

u=1.27m/s

Question Parameters:

a slight push toward an end stop 1.80 meters away

he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2

Generally the equation for the  third equation of motion    is mathematically given as

Vf^2 = Vi^2 + 2ad

Therefore

0=u^2+0.45*1.8

u=1.27m/s

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Answer:

wegkwe fhkrbhefdb

Explanation:B

Explanation:

Those tools that helps to make our work easier ,faster and more convenient in our daily life it is called simple Machine.

Answer:

simultaneously

Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

The given parameters include:

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]

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A

Explanation:

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