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2 years ago

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The wall of a large room is covered with acoustic tile in which small holes are drilled 4.6 mm from center to center. How far ca

n a person be from such a tile in meters and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be 4.0 mm, and the wavelength of the room light to be 550 nm?

1 answer:

Zielflug [23.3K]2 years ago

8 0

Answer:

L= 27.42m

Explanation:

We have the variables

$D=4.6*10^{-3}m$

$d= 4*10^{-3}m$

$\lambda = 550*10^{-9}m$

Use the relation,

$L=\frac{D}{\theta_R} = \frac{D_d}{1.22\lambda}$

$L= \frac{(4.6*10^{-3}m)(4*10^{-3}m)}{1.22(550*10^{-9}m)}$

$L= 27.42m$

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To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.

lozanna [386]

Answer:

a)

Explanation:

A block sliding down an inclined plane, is subject to two external forces along the slide.
One is the component of gravity (the weight) parallel to the incline.
If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

$F_{gp} = m*g* sin \theta (1)$

(taking as positive the direction of the movement of the block)

The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

$F_{f} = \mu_{k} * N (2)$

The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

$N = m*g* cos \theta (3)$

In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

$m*g* sin \theta = \mu_{k} * m*g* cos \theta$

As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

$\mu_{k} = tg \theta$

8 0

1 year ago

|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r

Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

$F = ma$

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

$a = \frac{F}{m}$

$a = \frac{11*10^3}{80}$

$a = 137.5m/s$

From the cinematic equations of motion we know that

$v_f^2-v_i^2 = 2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

$v_f^2-v_i^2 = 2ax$

$0-v_i^2 = 2(-g)x$

$v_i =\sqrt{2gx}$

$v_i = \sqrt{2*9.8*4.8}$

$v_i = 9.69m/s$

From the equation of motion where acceleration is equal to the velocity in function of time we have

$a = \frac{v_i}{t}$

$t = \frac{v_i}{a}$

$t =\frac{9.69}{137.5}$

$t = 0.0705s$

Therefore the time required is 0.0705s

4 0

2 years ago

Read 2 more answers

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

7 0

2 years ago

Come si compongono due forze che agiscono in diversi punti di un corpo rigido? Oof

bagirrra123 [75]

Answer:

Explanation:

I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

7 0

1 year ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

1 year ago