Ask question

Ask question

All categories

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 $1.904\ rad/s^2$ .

You might be interested in

Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1

7nadin3 [17]

Answer:

A). σ = 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B). $\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C). $U=10.322$ J

Explanation:

A). We know magnitude of charge per unit area for a conducting plate is given by

$\sigma =k.\varepsilon _{0}.E$

where, E is resultant electric field = 1.2 x $10^{6}$ V/m

$\varepsilon _{0}$ is permittivity of free space = 8.85 x $10^{-12}$ $C^{2}$ /N- $m^{2}$

k is dielectric constant = 3.6

∴ $\sigma =k.\varepsilon _{0}.E$

= 3.6 x 8.85 x $10^{-12}$ x 1.2 x $10^{6}$

= 3.823 x $10^{-5}$ $C^{2}$ /N- $m^{2}$

B).Now we know that the magnitude of charge per unit area on the surface of the dielectric plate is given by

$\sigma ^{'}=\sigma\left ( 1-\frac{1}{k} \right )$

$\sigma ^{'}=3.823\times 10^{-5}\left ( 1-\frac{1}{3.6} \right )$

$\sigma ^{'}=2.76\times 10^{-5}$ C/ $m^{2}$

C).

Area of the plate, A = 2.5 $cm^{2}$

= 2.5 x $10^{-4}$ $m^{2}$

diameter of the plate, d = 1.8 mm

= 1800 m

∴ Total energy stored in the capacitor

$U=\frac{1}{2}k\varepsilon _{0}E^{2}Ad$

$U=\frac{1}{2}\times 3.6\times8.85 \times10^{-12}\times\left ( 1.2\times 10^{6} \right ) ^{2}\times 2.5\times 10^{-4}\times 1800$

$U=10.322$ J

4 0

2 years ago

Suppose that the current in the solenoid is i(t. within the solenoid, but far from its ends, what is the magnetic field b(t due

Mkey [24]

The answer is B(t) = constants x I(t)

Please take precaution on the point that it is an independent field of its radial position, if the point is measured well in the solenoid. (also the radial position is the axis of its symmetry)

7 0

2 years ago

Read 2 more answers

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

5 0

2 years ago

The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o

MrMuchimi

Answer:

The total kinetic energy of both particles is $2.43\times10^{-13}$

Explanation:

Given that,

Kinetic energy of nucleus $K.E= 1.24\times10^{-13}\ J$

Kinetic energy of proton $K.E= 2.47\times10^{-13}\ J$

Radius of proton $r= 0.9\times10^{-15}\ m$

We need to calculate the final potential energy

Using formula of final potential energy

$U=\dfrac{kq^2}{r}$

Put the value into the formula

$U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}$

$U_{f}=1.28\times10^{-13}\ J$

We need to calculate the initial energy of both the particles

Using formula of energy

$E_{i}=(K.E_{n}+K.E_{p})+U_{i}$

$E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0$

$E_{i}=3.71\times10^{-13}\ J$

We need to calculate the total kinetic energy of both particles

Using conservation of energy

$E_{i}=E_{f}$

$E_{i}=K.E_{f}+U_{f}$

$3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}$

$K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}$

$K.E_{f}=2.43\times10^{-13}$

Hence, The total kinetic energy of both particles is $2.43\times10^{-13}$

3 0

2 years ago

Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear

dezoksy [38]

Answer:

Explanation:

Wheel completes four revolution.

The linear displacement is zero.

The angular displacement is 4 x 2π = 8π radian.

So, option (c) is correct.

4 0

2 years ago