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2 years ago

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An airplane is flying at a constant velocity through the air. What is the relationship between the magnitudes of the four forces

shown in the diagram?
A. L = W; F < D
B. L = W; F = D
C. L = W = F = D
D. L = W; F > D

1 answer:

zalisa [80]2 years ago

5 0

Answer:

(B) L=W; F=D

Explanation:

Since the airplane's velocity is constant, the airplane is not accelerating in any direction. In absence of acceleration, all forces acting on the plane are balanced so that the net forces in the horizontal and the vertical directions are zero. Given the drawing, this implies that

Forward force = Drag force (F=D)

Weight = Lift (W=L)

which corresponds to option (B)

(it is not C because nothing in the text says that the vertical and horizontal components have to have same magnitude)

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A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet an

kykrilka [37]

Answer:

a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

$m_Hv_H + m_Av_A = m_Hv_2+m_Av_2$

From the question given, we can deduce that the anvil is at rest;

∴ $v_A = 0$ ; then, we have:

$m_Hv_H + 0 = (m_H+m_A) v_2$

Making $v_2$ the subject of the formula; we have:

$v_2 = \frac{m_Hv_H}{m_H + m_A}$ ------- Equation (1)

Also, from the second diagram; there is a representation of a free body momentum of the hammer head;

From the diagram;

F = impulsive force exerted on the rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

$m_Hv_H - F \delta t = m_Hv_2$

$- F \delta t = - m_Hv_H + m_Hv_2$

$F \delta t = m_Hv_H - m_Hv_2$

$F \delta t = m_H(v_H - v_2)$ ------- Equation (2)

Using the function of the kinetic energy of the hammer before impact $T_1$ ; we have:

$T_1 = \frac{1}{2} m_Hv_H^2$ -------- Equation (3)

We determine the mass of the hammer $m_H$ by using the formula from:

$W_H = m_Hg$

where;

$W_H$ = weight of the hammer

$m_H$ = mass of the hammer

g = acceleration due to gravity

Making $m_H$ the subject of the formula; we have:

$m_H = \frac{W_H}{g}$

$m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}$

$m_H = 0.04658 \ lb.s^2/ft$

Now;

$T_1 = \frac{1}{2} m_Hv_H^2$

$T_1 = \frac{1}{2}*(0.04658 \ lb.s^2 /ft) *(20 \ ft/s)^2$

$T_1 = \frac{18.632 }{2}$

$T_1 = 9.316 \ ft.lb$

After the impact $T_2$ ; the final kinetic energy of the hammer and anvil can be written as:

$T_2 = \frac{1}{2}(m_H +m_A)v^2_2$

Recall from equation (1) ; where $v_2 = (\frac{m_Hv_H}{m_H+m_A})$ ; if we slot that into the above equation; we have:

$T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2$

$T_2 = \frac{1}{2} \frac{m^2_H +v^2}{m_H+m_A}$

$T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})$

Also; from equation (3)

$T_1 = \frac{1}{2} m_Hv_H^2$ ; Therefore;

$T_2 = T_1 (\frac{m_H}{m_H+m_A})$ ----- Equation (4)

a)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil $m_A = \frac{W_A}{g}$

where we replace; $W_A \ with \ \infty$ and g = 32.2 ft/s²

$m_A = \frac{\infty}{32.2 \ ft/s}$

However ; from equation (1)

$v_2 = \frac{m_H v_H}{m_H + m_A}$

$v_2 = \frac{0.04658*20}{0.04658+ \ \infty}$

$v_2 = 0$

From equation (2)

$F \delta t = m_H(v_H + v_2)$

$F \delta t = (0.04658 lb .s^2 /ft )(20ft/s - 0)$

$F \delta t = \ 0.932 \ lb.s$

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

For the energy absorbed by the rivet ; we have:

$T_2 = T_1 (\frac{m_H}{m_H+m_A} )$

where;

$T_1= 9.316 \ ft.lb$

$m_H = 0.04658 \ lb.s^2/ft$

$m_A = \infty$

Then;

$T_2 = (9.316 \ ft.lb) (\frac{0.04658\ lb.s^2/ft)}{0.04658 \ lb.s^2/ft+ \infty} )$

$T_2 = (9.316 \ ft.lb)* 0$

$T_2 = 0$

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = $T_1 - T_2$

ΔT = 9.316 ft.lb - 0

ΔT ≅ 9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow when the anvil has an infinite mass support = 9.32 ft.lb

b)

Due to the broadness of this question, the text is more than 5000 characters, so i was unable to submit it after typing it . In the bid to curb that ; i create a document for the answer for the part b of this question.

The attached file can be found below.

5 0

2 years ago

Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s

AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

sin ( Q ) = m*λ /d

m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

r = sqrt(2.8^ + 0.0025^) = 2.8

sin(Q) = 0.0025 / 2.8

Hence. m = 0.00085*0.0025 / 2.8*(600*10^-9)

m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer: Phase difference = pi/4 radians

6 0

2 years ago