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A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer 1
Answer:

Answer:

The answer is below

Explanation:

a) Using the formula:

\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2

b)

\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev

Related Questions

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Which of the following types of waves is not part of the electromagnetic spectrum? A) microwaves B) gamma rays C) ultraviolet radiation D) radio waves E) sound waves
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If the radio waves transmitted by a radio station have a frequency of 83.5 MHz, what is the wavelength of the waves, in meters

Answers

Answer: wavelength =3.52m

Explanation:

,λ=c/μ

where c=speed of the light,λ=wave length, μ=frequncy

c=3x10^8m/s

And

μ=83.5/MHz =85.3x10^6Hz==85.3x10^6Hz=

=85.3x10^6s-1

λ=c/μ

=3x10^8m/s/85.3x10^6s-1

=3.51699883

=3.52m

A 1500 kg car moving at 25 m/s hits an initially uncompressed horizontal ideal spring with spring constant (force constant) of 2.0 × 106 N/m. What is the maximum distance the spring compresses?

Answers

Answer:

x = 0.68 meters

Explanation:

It is given that,

Mass of the car, m = 1500 kg

Speed of the car, v = 25 m/s

Spring constant of the spring, k=2* 10^6\ N/m

When the car hits the uncompressed horizontal ideal spring the kinetic energy of the car is converted to the potential energy of the spring. Let x is the maximum distance compressed by the spring such that,

(1)/(2)mv^2=(1)/(2)kx^2

x=\sqrt{(mv^2)/(k)}

x=\sqrt{(1500* (25)^2)/(2* 10^6)}

x = 0.68 meters

So, the spring is compressed by a distance of 0.68 meters. Hence, this is the required solution.

Final answer:

The maximum distance the spring compresses when a 1500 kg car moving at 25 m/s hits it, given a spring constant of 2.0 × 10⁶N/m, is approximately 0.53 meters or 53 centimeters.

Explanation:

In this specific problem, we can apply the conservation of energy principle, where the initial kinetic energy of the car is converted into potential energy stored in the spring when the car comes to a stop. The formula for kinetic energy is K = 1/2 × m× v² and for potential energy stored in a spring is U = 1/2×k × x², where m = mass of the car, v = velocity of the car, k = spring constant, and x = maximum distance the spring is compressed.

By setting the kinetic energy equal to potential energy (since no energy is lost), we get 1/2 × m×v² = 1/2×k×x². Solving this equation for x (maximum compression of the spring), we obtain x = sqrt((m×v²)/k). Substituting the given values, x = sqrt((1500 kg× (25 m/s)²) / (2.0 × 10⁶ N/m)), which yields approximately 0.53 meters or 53 centimeters. Therefore, the maximum distance the spring compresses is 53 cm.

Learn more about Conservation of Energy here:

brainly.com/question/35373077

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Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009
The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
Hope i helped
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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the

Answers

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)

k = 1.4

T_2 = T_1((P_2)/(P_1))^{(k-1 )/(k)}\n\nT_2 = 1200((80)/(150))^{(1.4-1 )/(1.4)}\n\nT_2 = 1002.714K

Work done is given as;

W = (1)/(2) *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{(2W)/(m) } = √(2*C_p(T_1-T_2)) \n\nv_e = √(2*1004(1200-1002.714))\n\nv_e = √(396150.288) \n\nv_e = 629.41 \ m/s

Therefore, the exit velocity is 629.41 m/s

Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?

Answers

Answer:

The truck will reach there in 250 seconds.

Explanation:

The frequency due to doppler effect, when the observer is stationary and the source is moving towards it is

f_(obv)=(v)/(v-v_(s) ) f

where v= velocity of sound in air

v_(s)= velocity of source of sound

f= frequency of sound and

f_(obv)= frequency oberved due to Doppler effect

(v)/(v-v_(0) ) f = 460------------------------------------------( 1 )

The frequency due to doppler effect, when the observer is stationary and the source is moving away from it

f_(obv)=(v)/(v+v_(s) ) f

where v= velocity of sound in air

v_(s)= velocity of source of sound

f= frequency of sound and

f_(obv)= frequency oberved due to Doppler effect

(v)/(v+v_(0) ) f = 410-------------------------------------------( 2 )

Dividing ( 1 ) by ( 2 )

(v+v_(s) )/(v-v_(s) ) =(460)/(410)

(v+v_(s) )/(v-v_(s) ) =(46)/(41)

41v + 41v_(s) = 46v  - 46v_(s)

87v_(s)= 5v

v_(s)=(5)/(87)v

Velocity of Sound (v)= 348 m/s

v_(s)=20 m/s

Therefore, the truck is moving at 20 m/s.

Time=(Distance)/(Time)

Distance= 5000 m

Time=(5000)/(20)

Time= 250 s

Time = 4 min 10 sec          

         

         

         

         

We showed that the length of the pendulum of period 2.000 seconds on the Earth’s surface was 0.99396 meters. What period would this same pendulum have on the surface of Mars? What length would the pendulum be in order to have a period of 2.000 seconds?

Answers

To solve this problem it is necessary to apply the concepts related to the Period based on gravity and length.

Mathematically this concept can be expressed as

T= 2\pi \sqrt{(l)/(g)}

Where,

l = Length

g = Gravitational acceleration

First we will find the period that with the characteristics presented can be given on Mars and then we can find the length of the pendulum at the desired time.

The period on Mars with the given length of 0.99396m and the gravity of the moon (approximately 1.62m / s ^ 2) will be

T= 2\pi \sqrt{(l)/(g)}

T= 2\pi \sqrt{(0.99396)/(1.62)}

T = 4.921seg

For the second question posed, it would be to find the length so that the period is 2 seconds, that is:

T= 2\pi \sqrt{(l)/(g)}

2= 2\pi \sqrt{(l)/(1.62)}

l = 0.16414m

Therefore, we can observe also that the shorter distance would be the period compared to the first result given.
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