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3 years ago

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r

otating at 11.0 rev/s; 30.0 revolutions later, its angular speed is 18.0 rev/s. Calculate (a) the angular acceleration (rev/s2), (b) the time required to complete the 30.0 revolutions, (c) the time required to reach the 11.0 rev/s angular speed, and (d) the number of revolutions from rest until the time the disk reaches the 11.0 rev/s angular speed.

2 answers:

aivan3 [116]3 years ago

5 0

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

Elodia [21]3 years ago

3 0

Explanation:

Below is an attachment containing the solution

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A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C

GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0

3 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

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4 years ago

A circular flat coil that has N turns, encloses an area A, and carries a current i has its central axis parallel to a uniform ma

bogdanovich [222]

Answer:

Torque on the coil will be ZERO

Explanation:

As we know that the magnetic moment of the closed current carrying coil is always along its axis and it is given as

$M = N i A$

now we know that magnetic field is also along the axis of the coil so here as we know the equation of torque given as

$\tau = \vec M \times \vec B$

so we have

$\tau = M B sin0$

$\tau = 0$

5 0

4 years ago

Calculate the Reynolds number for an oil gusher that shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter

Alenkasestr [34]

Answer:

$Re=1992.24$

Explanation:

Given:

vertical height of oil coming out of pipe, $h=25\ m$

diameter of pipe, $d=0.1\ m$

length of pipe, $l=50\ m$

density of oil, $\rho = 900\ kg.m^{-3}$

viscosity of oil, $\mu=1\ Pa.s$

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

<u>Using the equation of motion:</u>

$v^2=u^2-2gh$

where:

v = final velocity

u = initial velocity

Putting the respective values:

$0^2=u^2-2\times 9.8\times 25$

$u=22.136\ m.s^{-1}$

<u>For Reynold's no. we have the relation as:</u>

$Re=\frac{\rho.u.d}{\mu}$

$Re=\frac{900\times 22.136\times 0.1}{1}$

$Re=1992.24$

6 0

3 years ago

A 500 kg car is at rest at the top of a 72 m high hill. The car rolls to the bottom of the hill. At the bottom of the hill, the

rjkz [21]

Explanation: Solution

1.

Gravitational potential energy

U=mgh=500*9.8*50

U=245000 J

2.

Kinetic energy is present at bottom of the hill

K=(1/2)mV2=(1/2)*500*27.82

K=193210 J

3.

Work done by friction

W=193210-245000=-51790 J

3 0

3 years ago