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4 years ago

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The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,

and AD = 15 lb. determine the force and moment acting at the trunk base point o. neglect the weight of the tree.

1 answer:

KATRIN_1 [288]4 years ago

3 0

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

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a freight train travels at v=60(1-e^-t) ft/s, where t is the elapsed time in seconds. Determine the distance traeled in tree sec

lesantik [10]

Answer

given,

$v = 60(1-e^{-t})\ ft/s$

t = 3 s

we know,

$v = \dfrac{dx}{dt}$

$a = \dfrac{dv}{dt}$

position of the particle

$dx = v dt$

integrating both side

$\int dx =\int 60(1-e^{-t}) dt$

$x = 60 t + 60 e^{-t}$

Position of the particle at t= 3 s

$x = 60\times 3+ 60 e^{-3}$

x = 182.98 ft

Distance traveled by the particle in 3 s is equal to 182.98 ft

now, particle’s acceleration

$a = \dfrac{dv}{dt}$

$a = \dfrac{d}{dt}60(1-e^{-t})$

$a = 60 e^{-t}$

at t= 3 s

$a = 60 e^{-3}$

a = 2.98 ft/s²

acceleration of the particle is equal to 2.98 ft/s²

7 0

4 years ago

X-rays with an energy of 86 keV scatter off of stationary electrons. How much energy do the photons scattering at 60° have? A. 7

AveGali [126]

Answer:

A. 79.3 keV

Explanation:

Because the procedure involves many steps for its resolution and it works faster on paper and pencil, the detailed solution of this exercise is attached as a scanned image of the procedure for review.

In the procedure, the initial values of the problem and the replacement of these values with the correct formulas for this process are taken into account.

7 0

3 years ago

A bicycle travels 6.10 km due east in 0.210 h, then 11.30 km at 15.0° east of north in 0.560 h, and finally another 6.10 km due

Virty [35]

Answer:

Explanation:

Given

First bicycle travels 6.10 km due to east in 0.21 h

Suppose its position vector is $r_1$

$r_1=6.10\hat{i}$

After that it travels 11.30 km at $15^{\circ}$ east of north in 0.560 h

suppose its position vector is $r_2$

$r_{2}=11.30\left ( cos15\hat{j}+sin15\hat{i}\right )$

after that he finally travel 6.10 km due to east in 0.21 h

suppose its position vector is $r_3$

$r_{3}=6.10\hat{i}$

so position of final position is given by

$r=r_1+r_{2}+r_{3}$

$\vec{r}=15.12\hat{i}+10.91\hat{j}$

$\vec{v_{avg}}=\frac{\vec{r}}{t}$

t=0.21+0.56+0.21=0.98 h

$\vec{v_{avg}}=15.42\hat{i}+11.13\hat{j}$

$|v_{avg}|=\sqrt{361.71}=19.01 km/hr$

For direction

$tan\theta =\frac{11.13}{15.42}=0.721$

$\theta =35.791^{\circ}$ w.r.t to x axis

5 0

4 years ago

An electron is released from rest at a distance of 0.570 m from a large insulating sheet of charge that has uniform surface char

Lelu [443]

Explanation:

Formula to calculate the electric field of the sheet is as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

And, expression for magnitude of force exerted on the electron is as follows.

F = Eq

So, work done by the force on electron is as follows.

W = Fs

where, s = distance of electron from its initial position

= (0.570 - 0.06) m

= 0.51 m

First, we will calculate the electric field as follows.

E = $\frac{\sigma}{2 \epsilon_{o}}$

= $\frac{4.60 \times 10^{-12}C/m^{2}}{2 \times 8.854 \times 10^{-12}C^{2}/N m^{2}}$

= 0.259 N/C

Now, force will be calculated as follows.

F = Eq

= $0.259 N/C \times 1.6 \times 10^{-19} C$

= $0.415 \times 10^{-19} N$

Now, work done will be as follows.

W = Fs

= $0.415 \times 10^{-19} N \times 0.51 m$

= $2.12 \times 10^{-20} J$

Thus, we can conclude that work done on the electron by the electric field of the sheet is $2.12 \times 10^{-20} J$ .

3 0

4 years ago

Read 2 more answers

The battery capacity of a lithium ion battery in a digital music player is 750 mA-h. The manufacturer claims that the player can

Oksi-84 [34.3K]

Answer:

The number of electrons is $6.3\times10^{21}\ electrons$

(D) is correct option.

Explanation:

Given that,

Battery capacity = 750 mA-h

Time t= 8 hours

Time t'=3 hours

We need to calculate the battery capacity

$Battery\ capacity=750\times10^{-3}\times3600$

$Battery\ capacity=2700\ A-s$

We need to calculate the number of electrons in 1 C Li

Using formula for number of electron

$n=\dfrac{1}{e}$

$n=\dfrac{1}{1.6\times10^{-19}}$

$n=6.25\times10^{18}\ electrons$

We need to calculate the number of electron in 2700 C

$2700\ C=2700\times6.25\times10^{18}=1.68\times10^{22}\ electrons$

The total number of electrons battery can deliver in 8 hours

$n=1.68\times10^{22}\ electrons$

We need to calculate the number of electron in 3 hours

Using formula of number of electrons

$n=\dfrac{n\times t'}{t}$

Put the value into the formula

$n=\dfrac{1.68\times10^{22}\times3}{8}$

$n=6.3\times10^{21}\ electrons$

Hence, The number of electrons is $6.3\times10^{21}\ electrons$

8 0

4 years ago

Read 2 more answers