PHYSICS COLLEGE

In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer 1

Answer:

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

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Bryan is riding his scateboard at 2.0 m/s. He accelerates at 0.5 m/s for 4.0 seconds. Calculate his final velocity.

Answers

Answer:

4 m/s^2

Explanation:

(0.5m/s * 4)+2 m/s

Modern wind turbines are larger than they appear, and despite their apparently lazy motion, the speed of the blades tips can be quite high-many times higher than the wind speed. A turbine has blades 56 m long that spin at 13 rpm .A. At the tip of a blade, what is the speed?
B. At the tip of a blade, what is the centripetal acceleration?
C. A big dog has a torso that is approximately circular, with a radius of 16cm . At the midpoint of a shake, the dog's fur is moving at a remarkable 2.5m/s .
D. What force is required to keep a 10 mg water droplet moving in this circular arc?
E. What is the ratio of this force to the weight of a droplet?

Answers

A) The speed of the tip of the blade is 76.2 m/s

B) The centripetal acceleration of the tip of the blade is $103.7 m/s^2$

D) The force required to keep the droplet moving in circular motion is 0.39 N

E) The ratio of the force to the weight of the droplet is 4.0

Explanation:

We know that the blade of the turbine is rotating at an angular speed of

$\omega = 13 rpm$

First, we have to convert this angular speed into radians per second. Keeping in mind that

$1 rev = 2 \pi$

$1 min = 60 s$

We get

$\omega = 13 rpm \cdot (2\pi rad/rev)/(60 s/min)=1.36 rad/s$

The linear speed of a point on the blade is given by:

$v=\omega r$

where

$\omega=1.36 rad/s$ is the angular speed

r is the distance of the point from the axis of rotation

For a point at the tip of the blade,

r = 56 m

Therefore, its speed is

$v=(1.36)(56)=76.2 m/s$

The centripetal acceleration of a point in uniform circular motion is given by

$a=(v^2)/(r)$

where

v is the linear speed

r is the distance of the point from the axis of rotation

In this problem, for the tip of the blade we have:

v = 76. 2 m/s is the speed

r = 56 m is the distance from the axis of rotation

Substituting, we find the centripetal acceleration:

$a=((76.2)^2)/(56)=103.7 m/s^2$

The force required to keep the 10 mg water droplet in circular motion on the dog's fur is equal to the centripetal force experienced by the droplet, therefore:

$F=m(v^2)/(r)$

where

m is the mass of the droplet

v is the linear speed

r is the distance from the centre of rotation

The data in this problem are

m = 10 mg = 0.010 kg is the mass of the droplet

v = 2.5 m/s is the linear speed

r = 16 cm = 0.16 m is the radius of the circular path

Substituting,

$F=(0.010)(2.5^2)/(0.16)=0.39 N$

The weight of the droplet is given by

$F_g = mg$

where

m = 10 mg = 0.010 kg is the mass of the droplet

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_g = (0.010)(9.8)=0.098 N$

The force that keeps the droplet in circular motion instead is

F = 0.39 N

Therefore, the ratio between the two forces is

$(F)/(F_g)=(0.39)/(0.098)=4.0$

Learn more about circular motion:

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A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

$\triangle m = (2d)/(\lambda) (n - 1)$

where

d = $3.95 * 10^(-2)m$

$\lambda = 400 * 10^(-9)m$

n = 1.00028

Now placing these values to the above formula

So, the number of bright-dark-bright fringe shifts observed is

$= (2 *3.95 * 10^(-2)m)/(400 * 10^(-9)m) (1.00028 - 1)$

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank.a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

Answers

Answer:

Explanation:

For refraction through a curved surface , the formula is as follows

μ₂ / v - μ₁ / u = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m

u , object distance = - .465 m

1 / v + 1.333 / .465 = (1 -1.333 )/1.95

1 / v + 2.8667 = - .171

1 / v = - 2.8667 - .171 = - 3.0377

v = - .3292 m

= - 32.92 cm

image will be formed in water.

c ) magnification = μ₁v / μ₂u , μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .

= (1.33 x 32.92) / (1 x 46.5)

= .94

size of image of teeth = .94 x 5

= 4.7 cm .

A positive magnification means the image is inverted compared to the object

Answers

False

Explanation:

A positive magnification means the image is erect compared to the object. Magnifications with values greater than one represent images that are smaller than the object. A magnification of 1 (plus or minus) means that the image is the same size as the object. If m has a magnitude greater than 1 the image is larger than the object, and an m with a magnitude less than 1 means the image is smaller than the object. If the magnification is positive, the image is upright compared to the object; if m is negative, the image is inverted compared to the object.

A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

Answers

Answer:

(a) Fw = 101.01 N

(b) W = 282.82 J

(d) N = 368.61 N

(e) Net force = 0 N

Explanation:

(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:

$F_g-F_w=0$ (1)