Ask question

Ask question

All categories

4 years ago

13

A motorcycle accelerates uniformly from rest and reaches a linear speed of 24.8 m/s in a time of 9.87 s. The radius of each tire

is 0.287 m. What is the magnitude of the angular acceleration of each tire

2 answers:

Vinvika [58]4 years ago

6 0

Answer:

8.756 rad/s²

Explanation:

Given that:

A motorcycle accelerates uniformly from rest, then initial velocity v_i = 0 m/s

It final velocity v_f = 24.8 m/s

time (t) = 9.87 s

radius (r) of each tire = 0.287 m

Firstly; the linear acceleration of the motor cycle is determined as follows:

$a_T$ =(V_f - v_i)/t

=(24.8-0)/9.87

=2.513 m/s²

Then; the magnitude of angular acceleration

α = $a_T$ /r

=2.513/0.287

=8.756 rad/s²

geniusboy [140]4 years ago

4 0

<h2>Answer:</h2>

8.75rad/s²

<h2>Explanation:</h2>

The tires of the motorcycle undergo a rolling motion. Therefore, the tangential acceleration, $a_{T}$ , of the tires is equal to their linear acceleration, a. i.e

$a_{T}$ = a --------------(i)

But, the tangential acceleration, $a_{T}$ , is the product of the angular acceleration, $\alpha$ , and the radius of the each of the tires. i.e

$a_{T}$ = r $\alpha$ ------------(ii)

<em>Combine equations (i) and (ii) as follows;</em>

a = r $\alpha$ --------------(iii)

<em>Also, the linear acceleration, a, is given by;</em>

a = $\frac{v - u}{t}$ ------------------(iv)

Where;

v = final linear speed of the tire

u = initial linear speed of the tire

t = time taken for the motion

Combine equations iii and iv as follows;

$\frac{v - u}{t}$ = r $\alpha$ ------------------(v)

<em>From the question;</em>

v = 24.8m/s

u = 0 (since the motorcycle accelerates from rest)

t = 9.87s

r = 0.287m

<em>Substitute these values into equation (v) as follows;</em>

$\frac{24.8 - 0}{9.87}$ = 0.287 $\alpha$

$\frac{24.8}{9.87}$ = 0.287 $\alpha$

2.51 = 0.287 $\alpha$

$\alpha$ = $\frac{2.51}{0.287}$

$\alpha$ = 8.75rad/s²

Therefore, the angular acceleration of each tire is 8.75rad/s²

You might be interested in

For the following types radio waves infared radiation visible light ultraviolet radiation gamma radiation of electromagnetic rad

ddd [48]

Answer: Option 5.

Decreases increases increases.

Explanation:

Electromagnetic spectrum refers to differences in frequencies, wavelengths and photon energies of electromagnetic radiation.

From the top of the list to the bottom i.e from radio waves, infrared waves, visible lights, ultraviolet radiation, gamma radiation, wavelengths decreases, the frequency increases as the photon energy increases.

The electromagnetic radiations : radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays have high-frequency and short wavelengths. Their wavelengths decrease. The photon energy is directly proportional to wave frequency.

6 0

4 years ago

The position coordinate of a particle which is confined to move along a straight line is given by s =2t3−24t+6, where s is measu

Gre4nikov [31]

Answer:

a) the answer is t=4 seconds

b) acceleration is zero

c) displacement= 142 m

Explanation:

Given the position of the particle

$s=2t^3-24t+6$

a) the time required when velocity $v=72 m/s$

$v=72=\frac{ds}{st}=6t^2-24$

now we solve for time $t$

$6t^2=72+24=96\\\\\implies t^2=\frac{96}{6}\\\therefore t=4 s$

b) acceleration when $v=30 m/s$

acceleration is the time derivative of velocity i.e

$a=\frac{dv}{dt}=\frac{d}{dt}(30)=0$

c) the net displacement of the particle during the interval t = 1 s to t = 4 s is

$s_4-s_1=2t^3-24+6|_{t=4}-2t^3-24+6|_{t=1}\\s_4-s_1=126-(-16)=142 m$

7 0

4 years ago

Help me!!! I really need it

Alenkinab [10]

Answer:

The reading on the ammeter A₁ , should be 2 Amp.

Explanation:

Given circuit shows all the bulbs are connected in parallel and ammeter A(T) at the source reads 6 Amp.

So, as the bulbs are in parallel connection, the current gets divided equal to each bulb.

so, the reading on the ammeter A₁ , should be 2 Amp.

But it will show 6 Amp, which is three times of the required value(2 amp).

This is because there was a mistake while making the circuit connections.

Instead of making connections as given circuit, the connections were made different as uploaded circuit.

Instead of connecting ammeter A₁ to only one bulb, it is connected to all the bulbs.

To correct this, remove the wire connecting ammeter and the second bulb.

And connect the second bulb directly to the switch.

3 0

4 years ago

A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conducting rails separated by a distance d. The rod ha

Lilit [14]

Answer:

a) $\vec{d} =B-A$

b) $I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c) $v \approx 0.6m/s$

Explanation:

From the question we are told that

Magnetic field strength $B=3.6T$

Distance traveled $d=7m$

Mass $m=4.7 kg$

Resistance $r=8.2 ohm$

Gravitational acceleration $g=9.8m/s^2$

$\theta =90 \textdegree$ Because of perpendicularity

a)

Generally the direction of the current will be given as

$\vec{d} =B-A$

Because it opposes increases of magnetic flux

b)

Generally the equation for induced EMF $E$ is mathematically given as

$E=B*v*d*\sin \theta$

$E=B*v*d*\sin 90 \textdegree$

Generally the equation for induced current $I$ is mathematically given as

$I=E/R$

$I=\frac{B*v*d*\sin 90 \textdegree}{R}$

c)

Generally the the equation for force F at terminal speed is mathematically given as

$F=mg$

$mg=B*I*d*\sin 90 \textdegree$

$mg=B*(\frac{B * v * d }{R}) *d$

$v=\frac{m*g*R}{B^2*D^2}$

$v=\frac{4.7*9.8*8.2}{3.6^2*7^2}$

$v=0.59475m/s$

$v \approx 0.6m/s$

7 0

3 years ago

deff fn [24]

Answer:

31.28m/s

Explanation:

We can use the third key equation of accelerated motion

Δd = viΔt+1/2aΔt^2

subtitute in the values we know

41.6 = vi(1.89)+1/2(-9.8)(1.89)^2

41.6 = vi(1.89) - 17.52

59.12 = vi(1.89)

vi = 31.28

4 0

3 years ago