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Alenkasestr [34]

3 years ago

13

A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the c

oefficient of static friction between the floor and the ladder is 0.28. what distance, measured along the ladder from the bottom, can a 60-kg person climb before the ladder starts to slip?

2 answers:

Marizza181 [45]3 years ago

8 0

Answer:

d=1.18m

Explanation:

A 5.0-m long, 12-kg uniform ladder rests against a smooth vertical wall with the bottom of the ladder 3.0 m from the wall. the coefficient of static friction between the floor and the ladder is 0.28. what distance, measured along the ladder from the bottom, can a 60-kg person climb before the ladder starts to slip?

frictional force =uFn

u=coefficient of static friction

Fn=normal force

assuming forces in the y direction, we ave

EFy=0

Efy=Fn-Fg-Fp

Fn=Fg+Fp

Fn=120N+600N

Fn=720N

we recall tat Frctional force=uFn

0.28*720N

201.6N

summing te forces in te x direction

EFx=Fr-fWall

Fr=frictional force

Fwall=force in the wall

Fr=Fwall

Fwall=201.6N

the angle te ladder makes wit round will be

cos $\alpha$ =3/5

$\alpha$ =53.13

Next, we can turn to calculating the net torque

Choose the pivot point at the bottom of the ladder

(this choice eliminates FN and Fr).

rgrav =1/2*5=2.5

rwall=5m

rman=d

twall== r( wall)(Fwall)sinalpa

5*201.6sin53.13

twall=+806.39Nm

τ grav = r( grav )(Fgrav )sin53.13

τ grav=2.5*120*)sin53.13

-239.99nm

τ person = r( person )(Fperson )sin53.13

d*600Nsn53.13

-479.99d

∑τ = τ wall + τ grav + τ person

= 0

summation of te torque

te negative sin is for troques in te clockwise direction

+806.39Nm-239.99nm-479.99d=0

566.4=479.99

d=1.18m

Semmy [17]3 years ago

7 0

First establish the summation of the forces acting int the ladder

Forces in the x direction Fx = 0 = force of friction (Ff) – normal force in the wall(n2)

Forces in the y direction Fy =0 = normal force in floor (n1) – (12*9.81) –( 60*9.81)

So n1 = 706.32 N

Since Ff = un1 = 0.28*706.32 = 197,77 N = n2

Torque balance along the bottom of the ladder = 0 = n2(4 m) – (12*9.81*2.5 m) – (60*9.81 *x m)

X = 0.844 m

5/ 3 = h/ 0.844

H = 1.4 m can the 60 kg person climb berfore the ladder will slip

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The normal eye, myopic eye and old age

yanalaym [24]

Answer:

1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

$\frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}$

f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

1 / f = 1/15 + 1 / 1.5

1 / f = 0.733

f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

f’₀ / f = 1.5 / 1.36

f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

sin θ = y / f

where f is the distance of the lens (eye)

y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

y / f = 1.22 λ / D

y = 1.22 λ f / D

where D is the diameter of the eye

D = 2R₀

D = 2 0.1

D = 0.2 cm

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

\frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

$\frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}$

$\frac{1}{f}$ = 0.7066

f = 1.415 cm

therefore the diffraction is

y = 1.22 550 10⁻⁹ 1.415 / 0.2

y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

d = 2y

d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

$\frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}$

$\frac{1}{f}$ = 0.733

f = 1.36 cm

diffraction is

y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

y = 4.56 10-6 m

the diffraction diameter is

d_myope = 2y

d_myope = 9.16 10-6 m

$\frac{d_{normal}}{d_{myope}}$ = 9.49 /9.16

\frac{d_{normal}}{d_{myope}} = 1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

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3 years ago

A uniform plank of mass 10kg and length 10m rests on two supports, A and B as shown. A boy of weight 500N stands at a distance o

kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

Taking summation of the moment about point A, we get the following equilibrium equation: (taking clockwise direction as positive)

$W(2\ m) - U^2(7\ m) = 0$

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

$(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\$

<u>U² = 142.86 N</u>

Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

$W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\$

<u>U¹ = 357.14 N</u>

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Answer:

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2 years ago

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A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun

timurjin [86]

Answer:

a) $T=1.43\times 10^{-3}\ s$

b) $d=0.0429\ m$

c) $\lambda=0.4857\ m$

d) $f_o=767.7\ Hz$

Explanation:

Given:

velocity of the sound from the source, $S=340\ m.s^{-1}$

original frequency of sound from the source, $f_s=700\ Hz$

speed of the source, $v_s=30\ m.s^{-1}$

(a)

We know time period is inverse of frequency:

Mathematically:

$T=\frac{1}{f}$

$T=\frac{1}{700}$

$T=1.43\times 10^{-3}\ s$

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

$d=v_s.T$

$d=30\times 1.43\times 10^{-3}$

$d=0.0429\ m$

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

$\lambda=\frac{S}{f}$

$\lambda=\frac{340}{700}$

$\lambda=0.4857\ m$

(d)

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<u>According to the Doppler's effect:</u>

$\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s}$ ...........................(1)

where:

$f_o\ \&\ v_o$ are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

$\therefore v_o=0\ m.s^{-1}$

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$\frac{f_o}{700}= \frac{340+0}{340-30}$

$f_o=767.7\ Hz$

5 0

3 years ago

As of 2016, Nevada has a current minimum wage of $8.25 an hour ($/hr). If someone was really picky about how much they owed, how

xz_007 [3.2K]

Once again, you'd need to know that there are 60 seconds in a minute, and 60 minutes in an hour :)

I'd say converting the minimum wage into cents rather than dollars would make this problem a lot easier. $8.25 = 825 ¢.

So if this person is earning 825 ¢ in an hour, we should divide 825 by 60 to find out how much they're making in a minute:

825 ÷ 60 = 13.75 ¢

Now, we just need to divide by 60 again to work out how much that is in seconds:

13.75 ÷ 60 = 0.229 ¢

So to answer your question, this person would make 0.229 ¢ a second (¢/s) on the job with minimum wage. Converting this value to dollars wouldn't be viable (as it'd just be $0.00, so it's best to leave the answer in cents!)

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3 years ago

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