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brilliants [131]

2 years ago

12

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

tires rolled without slipping, what was the magnitude of the average angular acceleration of the tires during the time the car slowed to a rest

1 answer:

cupoosta [38]2 years ago

4 0

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

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The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th

Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em> $1.83\times 10^{7}\,kg$ <em>. The wheels of the locomotive push back on the tracks with a constant net force of </em> $7.50\times 10^{5}\,N$ <em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force ( $F$ ), measured in newtons, then acceleration ( $a$ ), measured in meters per square second, must be constant and can be found by the following expression:

$a = \frac{F}{m}$ (1)

Where $m$ is the mass of the freight train, measured in kilograms.

If we know that $F = 7.50\times 10^{5}\,N$ and $m = 1.83\times 10^{7}\,kg$ , then the acceleration experimented by the train is:

$a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}$

$a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$

Now, the time taken to accelerate the freight train from rest ( $t$ ), measured in seconds, is determined by the following formula:

$t = \frac{v-v_{o}}{a}$ (2)

Where:

$v$ - Final speed of the train, measured in meters per second.

$v_{o}$ - Initial speed of the train, measured in meters per second.

If we know that $a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}$ , $v_{o} = 0\,\frac{m}{s}$ and $v = 22.222\,\frac{m}{s}$ , the time taken by the freight train is:

$t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }$

$t = 542.265\,s$

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0

1 year ago

Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a

mote1985 [20]

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0

1 year ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

2 years ago

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

Gnesinka [82]

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

$U=mgh$

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

6 0

2 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

aleksandrvk [35]

Answer:

Part a)

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

$t = \frac{t_1}{5}$

Explanation:

Part a)

As we know that it is having constant torque so here the time taken by it to accelerate is given as

$\omega_f = \omega_i + \alpha t$

$\omega_1 = 0 + \alpha t_1$

$t_1 = \frac{\omega_1}{\alpha}$

Part b)

angular displacement is given as

$\theta = \omega_i t_1 + \frac{1}{2}(\alpha) t_1^2$

$\theta = 0 + \frac{1}{2}\alpha t_1^2$

$\theta = \frac{1}{2}\alpha t_1^2$

Part c)

As we know that the angular deceleration produced by the brakes is given as

$\alpha_d = - 5\alpha$

now we have

$\omega_f = \omega _i + \alpha t$

$0 = \omega_1 - 5 \alpha_1 t$

$t = \frac{\omega_1}{5 \alpha}$

As we know that

$t_1 = \frac{\omega_1}{\alpha}$

so we have

$t = \frac{t_1}{5}$

6 0

2 years ago