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garik1379 [7]
4 years ago
6

For a statically indeterminate axially loaded member, Group of answer choices The total deflection between end A and end B of an

axially loaded member must not be zero. The summation of the forces is not zero. The summation of the reaction forces is equal to the applied load. The applied load must exceed the total reaction forces. The compatibility condition cannot be satisfied, since it is indeterminate.
Engineering
1 answer:
blsea [12.9K]4 years ago
4 0

Answer: The summation of the reaction forces is equal to the applied load.

Explanation:

When solving an indeterminate structure, it is important for one to satisfy the force-displacement requirements, compatibility, and the equilibrium of the structure.

With regards to the above question, for a statically indeterminate axially loaded member, the summation of the reaction forces will be equal to applied load. Here, the summation of the reactive forces will then be zero. This is a condition that is necessary to solve the unknown reaction forces.

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A frying pan is connected to a 150-volt circuit. If the resistance of the frying pan is 25 ohms, how many amperes does the fryin
mario62 [17]

Answer:

Explanation:

Ohms Law I=E/R (resistive requires no power factor correction)

150/25= 6 amps

5 0
3 years ago
1.0•10^-10 standard form
Drupady [299]

Answer:

1.0 * 10^{-10} = 0.0000000001

Explanation:

Given

1.0 * 10^{-10}

Required

Convert to standard form

1.0 * 10^{-10}

From laws of indices

a^{-x} = \frac{1}{a^x}

So, 1.0 * 10^{-10} is equivalent to

1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}

1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}

1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}

1.0 * 10^{-10} = 1.0 * 0.0000000001

1.0 * 10^{-10} = 0.0000000001

Hence, the standard form of 1.0 * 10^{-10} is 0.0000000001

3 0
3 years ago
A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?
Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0
4 years ago
Transformation is the change of one energy form to another.<br> O True<br> O False
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2 years ago
Read 2 more answers
A girl operates a radio-controlled model car in a vacant parking lot. The girl's position is at the origin of the xy coordinate
lawyer [7]

Answer:

a) 17.20

b) 11.31

c) 14.42

d) 12.65

Explanation:

(a)

The girl is at the origin of the x,y coordinates  (i.e 0,0,0  )

the position vector of the car at time 't' secs is

\vec{r}= 2+2t^2, 6+t^3,0

at t=2s,  the position vector is

\vec{r}= 10, 14,0

Therefore, the the distance between the car and the girl is

s= \sqrt{(10-0)^2+(14-0)^2+(0-0)^2)}\

s = 17.20

(b)

The position of the car at  t = 0s is \vec{r}_0 = 2,6,0

The position of the car at t = 2s is \vec{r}_2 = 10,14,0

The distance of the car traveled in the interval from t=0s to t=2 s is as follows:

s_{02}= \sqrt{(10-2)^2+(14-6)^2+(0-0)^2)} \\ \\ s_{02}  = 11.31

(c)

The position vector of the car at time 't' secs is

\vec{r}= 2+2t^2, 6+t^3,0

The velocity of the car is

\vec{v}=\dfrac{d\vec{r}}{dt}= 4t, 3t^2,0

the direction of the car's velocity at t = 2s is  going to be

\vec{v}\mid _t=2 8, 12,0

Thus; The speed of the car is

v_{t=2}= \sqrt{8^2+12^2+0^2} \\ \\  v_{t=2}= 14.42

(d)    the car's acceleration is:

\vec{a}=\frac{d\vec{v}}{dt}= 4, 6t,0

The magnitude of car's acceleration at t=2s is

\mid \vec{a}\mid _{t=2}=\sqrt{4^2+12^2+0^2} \\ \\ \mid \vec{a}\mid _{t=2}= 12.65

5 0
3 years ago
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