Normally the rate at which you expend energy

In 1967, new zealander burt munro set the world record for an indian motorcycle, on the bonneville salt flats in utah, with a ma

Fed [463]

Draw a velocity-time diagram as shown below.

Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s

The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
= (1/2)*(6.705 m/s²)*(12.2446 s)²
= 502.64 m

Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m

3 0

3 years ago

shows a conical pendulum, in which the bob (the small object at the lower end of the cord) moves in a horizontal circle at const

Contact [7]

Answer:

a) $T=0.40 N$

b) $T=1.9 s$

Explanation:

Let's find the radius of the circumference first. We know that bob follows a circular path of circumference 0.94 m, it means that the perimeter is 0.94 m.

The perimeter of a circunference is:

$P=2\pi r=0.94$

$r=\frac{0.94}{2\pi}=0.15 m$

Now, we need to find the angle of the pendulum from vertical.

$tan(\alpha)=\frac{r}{L}=\frac{0.15}{0.90}=0.17$

$\alpha=9.44 ^{\circ}$

Let's apply Newton's second law to find the tension.

$\sum F=ma_{c}=m\omega^{2}r$

We use centripetal acceleration here, because we have a circular motion.

The vertical equation of motion will be:

$Tcos(\alpha)=mg$ (1)

The horizontal equation of motion will be:

$Tsin(\alpha)=m\omega^{2}r$ (2)

a) We can find T usinf the equation (1):

$T=\frac {mg}{cos(\alpha)}=\frac{0.04*9.81}{cos(9.44)}=0.40 N$

We can find the angular velocity (ω) from the equation (2):

$\omega=\sqrt{\frac{Tsin(\alpha)}{mr}}=3.31 rad/s$

b) We know that the period is T=2π/ω, therefore:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{3.31}=1.9 s$

I hope it helps you!

8 0

3 years ago

The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coeffi

Nezavi [6.7K]

Answer:

Explanation:

Let T be the tension in the cord.

Impulse by cord = change in momentum of block A .

T x 5s = 10 ( 2 -0) = 20

T = 4 poundal .

acceleration of block B = 2 / 5 = 0.4 m /s²

Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .

= 8 ( 32 + .4 ) = 259.2 poundal

Frictional force on block A = 259.2 - 4 = 255.2 poundal

μ x 10 x 32 = 255.2

320μ = 255.2

μ =0 .8 .

8 0

3 years ago

A pulley of radius 8.0 cm is connected to a motor that rotates at a rate 7000 rad s-1 and then decelerate uniformly at a rate of

zlopas [31]

Answer:

(a) α = - 1000 rad/s²

Negative sign represents deceleration.

(b) θ = 3581 rotations

(c) L = 1800 m

(d) a = - 80 m/s²

Explanation:

(a)

using First equation of motion for angular motion:

ωf = ωi + αt

where,

ωf = Final Angular Speed = 2000 rad/s

ωi = Initial Angular Speed = 7000 rad/s

α = Angular Acceleration = ?

t = time = 5 s

Therefore,

2000 rad/s = 7000 rad/s + α(5s)

α = (2000 rad/s - 7000 rad/s)/5 s

α = - 1000 rad/s²

Negative sign represents deceleration.

(b)

Using second equation of motion:

θ = ωi t + (1/2)αt²

where,

θ = No. of Rotations = ?

Therefore,

θ = (7000 rad/s)(5 s) + (1/2)(- 1000 rad/s²)(5 s)²

θ = 35000 rad - 12500 rad

θ = (22500 rad)(1 rotation/2π rad)

θ = 3581 rotations

(c)

Length of String = L = (Circumference of Pulley)(θ)

L = [2π(0.08 m)][3581 rotations]

L = 1800 m

(d)

Tangential Acceleration = a = rα

a = (0.08 m)(-1000 rad/s²)

a = - 80 m/s²

4 0

3 years ago

If one mile is 5280 feet, how many inches are in 3.94 km?

Yakvenalex [24]

Answer:

155 118 in

Explanation:

3.94 km * 1000 m / km * 39.37 in / m = 155 118 in

3 0

2 years ago