Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 13.9 revolutions/s. Find the linear speed v of the car, assuming that the tires are not slipping against the ground.

Complete Question

A system consists of a disk of mass 2.0 kg and radius 50 cm upon which is mounted an annular cylinder of mass 1.0 kg with inner radius 20 cm and outer radius 30 cm (see below). The system rotates about an axis through the center of the disk and annular cylinder at 10 rev/s.

(a) What is the moment of inertia of the system

(b) What is its rotational kinetic energy? axis 50 cm 30 cm 20 cm

Answer:

a) [tex]M.I=0.32kg.m^2[/tex]

b) [tex]K.E=621.8J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=2.0kg[/tex]

Disk Radius [tex]r_d=50cm=0.5m[/tex]

Mass of annular cylinder [tex]M_c=1.0kg[/tex]

Inner Radius of cylinder [tex]R_i=20cm=0.2m[/tex]

Outer Radius of cylinder [tex]R_o=0.3m[/tex]

Angular  Velocity [tex]\omega=10rev/s=2 \pi*10=62.83[/tex]

Generally the equation for moment of inertia is mathematically given by

[tex]M.I=0.5M r_d^2+0.5M_c(R_i^2+R_o^2)[/tex]

[tex]M.I=0.5(2*0.50)^2)+0.5*1*(0.2^2+0.30^)[/tex]

[tex]M.I=0.32kg.m^2[/tex]

Generally the equation for Rotational Kinetic Energy  is mathematically given by

[tex]K.E=0.5M.I \omega^2[/tex]

[tex]K.E=0.5 *0.32*62.83[/tex]

[tex]K.E=621.8J[/tex]

Explanation:

bosu

here is your answer

Solution :

The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]

So the distance appeared to Bob due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]

    [tex]$=7.1 \ Ly$[/tex]

Therefore, the time required to finish the race by Bob is

[tex]$t_b = \frac{L_b}{V_b}$[/tex]

  [tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]

So the distance appeared to Charlie due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]

    [tex]$=5 \ Ly$[/tex]

The time required to finish the race by Charlie is

[tex]$t_b = \frac{L_c}{V_c}$[/tex]

  [tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]

  = 5.77 year

Answer: 113.75

Explanation:

You know

acceleration = a = 3.5 m/s²

time = t = 5 seconds

initial velocity = u = 14 m/s

Unknown is distance = s = ?

Use equation: s = ut + [tex]\frac{1}{2}[/tex] at²

Substitute all the known values inside the equation:

s = (14*5) + 0.5 * 3.5 * 5²

s = 70 + 43.75 = 113.75 m

The car travels 113.75 metres.

Answer:

1.5 m is ur answer

Explanation:

Answer:

3.0−0.12=2.88 or 2.88÷0.04=72

0.04×3.0=0.12 and 0.04+3.0=3.04

Answer: 2.4 J

Explanation: Khan Academy

Answer:

Angle of reflection

Explanation:

The angle between the reflected ray and the normal line is called the angle of reflection

Answer:

Activity 2 Directions: Solve the following problems using Charles' Law,

1. A 3.8 mL of a gas in a container has a temperature of 2.0K. What would the volume be if the temperature becomes 8.0K?

2. If at 27°C the volume of a dry gas is 3.0 ml. What is the new temperature if the volume increased to 7.5 ml?

3. The temperature of the gas is 460,0 mL at 3.5°C. Calculate the new volume at 1.5°C

Answer:

mmmlbdhdjdkekkdnxnfjkckckcklclglglvkglglvkgkvkvkvkvkvkvkvkvkvkvkbkkbkbkkbbkkbkbkbkkbkbkblbkkbkb

Answer:

An Object is positively charged if it has more Positive Electrons in that object

Answer:

It will go to the right way 500N.

Explanation:

Because, up and down is 300 and 300 so it minus each other and get zero, that made the resultant force will never go up or down. And now we are focusing on the left and right. 900 is more than 400, so it will go to the right. But we also pull to the left too!! That make the resultant force is 500 to the right.

Answer:

a) By [tex]v^2 = u^2 + 2as[/tex] => a= 70291.70.

(b)By [tex]v = u + at[/tex] => t= 1.58 ms.

(c)By [tex]v^2 = u^2 - 2gh[/tex] => H = 46045.92 m.

Explanation:

a) By [tex]v^2 = u^2 + 2as[/tex]

[tex](950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70[/tex]

(b)By [tex]v = u + at[/tex]

[tex]950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms[/tex]

(c)By [tex]v^2 = u^2 - 2gh[/tex]

[tex]0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m[/tex]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

Answer:

B

Explanation:

because it take to the thought of a situation

Answer:

d i think

Explanation:

and are u using IXL???

The speed is in feet per seconds so change the length of the field from yards to feet.

1 yard = 3 feet

100 yards x 3 = 300 feet

The field is 300 feet long

Time = distance / speed

Time = 300 feet / 5 feet per second = 60 seconds = 1 minute

It will take 1 minute

Answer:

A person will take 60 Seconds to walk the distance of 100 yards.

Explanation:

Data Given ;

Distance ( d ) = 100 yards = 300 Ft

Speed ( v ) = 5 Ft/s

Time ( t ) = ?

The distance travelled in unit time is called speed.

formula ; [tex]v = \frac{d}{t} \\[/tex]

On putting values,

[tex]5 = \frac{300}{t}[/tex]

[tex]t =\frac{300}{5}[/tex]

[tex]t = 60 sec[/tex]

Hence the time taken by the person is 60 sec.

https://brainly.com/question/22610586?referrer=searchResults

#SPJ2

Answer:

[tex]{ \tt{initial \: velocity, \: u = 0}} \: (at \: rest) \\ { \tt{final \: velocity, \: v = 15 { {ms}^{ - 1} }}} \\ { \tt{time, \: t = 6s}} \\ { \bf{from \: first \: newtons \: equation \: of \: motion : }} \\ { \bf{v = u + at}} \\ { \tt{15 = 0 + (a \times 6)}} \\ { \tt{6a = 15}} \\ { \tt{acceleration, \: a = 2.5 \: {ms}^{ - 2} }}[/tex]

Answer:

Decrease occur in kinetic energy.

Explanation:

The kinetic energy decreases when the the Bob swings reaches to the maximum height because the motion of the bob slower down. At maximum height, the kinetic energy decreases whereas the value of potential energy is the highest. The main reason of higher potential energy is that it depends on height of an object while on the other hand, kinetic energy depends on the motion of an object so that's why the value of kinetic energy decreases and potential energy increases at maximum height of the bob.

Answer:

Distance

Explanation:

Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.

Mathematically, work done is given by the formula;

[tex] W = F * d[/tex]

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

A dependent variable is the event expected to change when an independent variable is manipulated.

Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.

F = 153 N

[tex]\theta = 11.3°[/tex]

Explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:

[tex]F_{x}[/tex] = 350 N - 200 N = 150 N

[tex]F_{y}[/tex] = 180 N - 150 N = 30 N

The magnitude of the resultant force F is given by

[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]

[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]

[tex]\:\:\:\:\:\:=153\:N[/tex]

To find the direction [tex]\theta[/tex], we use

[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]

or

[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]

Answer:

We know that the change in electric potential energy is defined as:

q*ΔV = ΔP

So, the change in the electric potential energy is the charge times the change in the electric potential.

For the case of an electron gas, we have:

q = -e

where -e is the charge of an electron (remember that is negative).

So, if the electron gains electric potential then:

ΔP > 0

this means that the final potential energy is larger than the initial one, then we have:

-e*ΔV > 0

This means that ΔV must be negative.

V₂ = electric potential at point 2, so it is the final electric potential

V₁ = electric potential at point 1, so it is the final electric potential

Then we should get:

ΔV = V₂ - V₁ < 0.

This means that:

V₂ < V₁

The correct option is b.

Answer:

[tex]I=0.902kg*m^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=92.0kg[/tex]

Radius [tex]r=0.140m[/tex]

Generally the equation for moment of Inertia is mathematically given by

 [tex]I = 0.5*m*r^2[/tex]

 [tex]I=0.5*(92)*0.14^2[/tex]

 [tex]I=0.902kg*m^2[/tex]

Answer:45,000

Explanation:

:TIL the average pencil holds enough graphite to draw a line about 35 miles long or to write roughly 45,000 words

Answer:

Explanation:

From the given information:

total mass = 25 kg

distance d = 100 m

height = 4.0 m

Force F = 9.0 N

The speed at (bottom) u = 5.0 m/s

Using the concept of energy conservation;

[tex]\dfrac{1}{2}mu^2 + W = \dfrac{1}{2}mv^2 + mgh[/tex]

divide both sides by m

[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]

multiply both sides by 2

[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]

[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]

[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]

Recall that:

W = Fd

∴

[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]

[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]

[tex]v^2 =(25.0 ) - 78.48 +72[/tex]

[tex]v^2 = 18.52 \ m^2/s^2[/tex]

[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]

v = 4.30 m/s

Answer:

a wheel and axle and a lever

The wheelbarrow has a wheel and the then the barrow. The wheel is an wheel and axle machine while the other part is a lever.

A simple machine is a device that is used to make work easier. It ensures that less effort is applied to overcome a large load.

The wheelbarrow has a wheel and the then the barrow. The wheel is an wheel and axle machine while the other part is a lever.

Learn more about simple machines:https://brainly.com/question/10075890

#SPJ2

Answer:

4 ohm

Explanation:

The equivalent resistance (Re) of three resistors in parallel is given by;

1/Re = 1/R1 + 1/R2 + 1/R3

Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm

1/Re= 1/16 + 1/16 + 1/8

1/Re= (0.0625) + (0.0625) + (0.125)

Re= 4 ohm

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

Kinetic friction is defined as the opposing force exerted by the surface on an object in contact with it, when there is relative motion between the two surfaces.

Here,

Mass of the box, m = 150 lb = 68.1 kg

Coefficient of kinetic friction, μ = 0.45

Maximum horizontal force that can be applied on the box is the kinetic frictional force. Frictional force,

F(k) = μmg

F(k) = 0.45 x 68.1 x 9.8

F(k) = 300.32 N

Now, the box sits on a ramp inclined at 60°

Coefficient of kinetic friction, μ = 0.45

The net force here acting on the box placed in the ramp is due to the kinetic frictional force and the weight of the box.

So,

Frictional force, F(k)' = μmgcosθ

F(k)' = 0.45 x M x 9.8 x cos 60

F(k)' = 2.2M

Weight of the box acting horizontally,

W = Mgsinθ

W = M x 9.8 x sin60

W = 8.5M

Therefore, net force,

Fn = W - F(k)'

Fn = 8.5M - 2.2M

Fn = 6.3M

The total force acting on the box is

F = F(k) - Fn

ma = 300.32 - 6.3M

Since, the box is moving with constant speed, the acceleration, a = 0

Therefore,

300.32 - 6.3M = 0

6.3M = 300.32

M = 300.32/6.3

M = 47.7 kg = 105.16 pound

Hence,

Maximum horizontal force that can be applied on the box is 300.32 N.

Mass of the heaviest box that can be pushed on the ramp at constant speed is 105.16 pound.

To learn more about kinetic friction, click:

https://brainly.com/question/29100737

#SPJ2

Answer:

Economists, I guess.

Explanation:

Answer:

225,000 kilometers per second

Explanation:

Have a nice day

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =[tex]F^>[/tex] × [tex]x^>[/tex]

W = [tex]Fxcos\theta[/tex]    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = [tex]Fxcos([/tex] 0° [tex])[/tex]

W = [tex]Fx[/tex] ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

[tex]W_{net[/tex] = ΔKE

= [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu²

we make F, the subject of formula

F = [tex]\frac{m}{2x}[/tex]( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N