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2 years ago

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The column has a mass of 600 lb/ft and a force of 50 kips applied at the top. The diagonal brace has a 10 kip axial force that c

an be either tension or compression (note that in seismic and wind loading on buildings, braces experience both tension and compression). Assume all elements are pin-connected at their ends, including the two supports.

1 answer:

drek231 [11]2 years ago

4 0

Answer: the maximum compression force = 59.47 kip

the minimum compression force = 43.33 kip

Explanation:

In this we required to do force balance in order to get maximum and minimum compression force in the column.

the picture below explains the steps to solve the question with diagrams to ease understanding.

You might be interested in

A 2kg object weighs 19.6N on earth. If the acceleration due to gravity on mars is 3.71 meters per second what is the object mass

Gre4nikov [31]

The acceleration due to gravity at the Earth's surface is 9.8 m/s/s

<span>So a 2 kg object will accelerate at 9.8 m/s/s if it is dropped..... actually all dropped objects will accelerate at the same rate. </span>

<span>The force of gravity on the object is given by Newton's 2nd Law </span>

<span>F = m . a </span>

<span>F = 2 . 9.8 = 19.6 N </span>

<span>Now go to Mars , with a lower gravitational field strength. It only accelerates falling objects at 3.71 m/s/s </span>

<span>So the force of gravity on the object is F = 2 . 3.71 = 7.42 N </span>

<span>But the answer is (2) 2 kg, because the mass of the object stays the same and that is what you are asked for. </span>
<span>Someone was trying to be tricky.</span>

3 0

3 years ago

The speed of the incoming water is 25 cm/s and the cross-sectional area of the hose carrying the water is 2.0 cm2. 1) Determine

Sindrei [870]

Answer:

The time taken is $t = 500 \ s$

Explanation:

From the question are told that

The speed of the incoming water is $v = 25 cm/s$

The cross-sectional area of the host pipe is $a = 2.0 cm^2$

Generally the rate at which the container is been filled is mathematically represented as

$\r V = v * a$

=> $\r V = 25 * 2$

=> $\r V = 50\ cm^3/s$

Generally $2.5 L = 25000 cm^3$

Generally the time taken is mathematically represented as

$t = \frac{25000}{50}$

=> $t = 500 \ s$

6 0

2 years ago

A 5.0-nC point charge is embedded at the center of a nonconducting sphere (radius = 2.0 cm) which has a charge of -8.0 nC distri

Igoryamba

Answer:

3.6 × 10⁵ N/C = 360 kN/C

Explanation:

Let R = 2.0 cm be the radius of the sphere and q = -8.0 nC be the charge in it. Let q₁ be the charge at radius r = 1.0 cm. Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πr³

q₁ = q(r/R)³. The electric field due to q₁ at r is E₁ = kq₁/r² = kq(r/R)³/r² = kqr/R³

The electric field due to the point charge q₂ = 5.0 nC is E₂ = kq₂/r².

So, the magnitude of the total electric field at r = 1.0 cm is

E = E₁ + E₂ = kqr/R³ + kq₂/r² = k(qr/R³ + q₂/r²)

E = 9 × 10⁹(-8 × 10⁻⁹ C × 1 × 10⁻² m/(2 × 10⁻² m)³ + 5 × 10⁻⁹ C/(1 × 10⁻² m)²)

E = 9 × 10⁹(-1 × 10⁻⁵ + 5 × 10⁻⁵)

E = 9 × 10⁹(4 × 10⁻⁵)

E = 36 × 10⁴ N/C = 3.6 × 10⁵ N/C = 360 kN/C

6 0

3 years ago

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the <em>electrical</em> force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

8 0

1 year ago

A computer hard disk 8.0 cm in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerat

dexar [7]

Answer:

a) 282.8 m/s

b) 13812 revolutions

Explanation:

a) The aceleration(a) is the angular velocity(w) divide by the time. The disk had cosated for 12 s in a constant angular velocity, so it will be:

a = w/t

610 = w/12

w = 7230 rad/s

The angular velocity is related by the linear velocity:

V = w*r, where r is the radius of the circumference ( r = 0.08/2 = 0.04 m)

V = 7230*0.04 = 292.8 m/s

b) The number of revolutions is the frequency (f) of the moviment, which is related to the angular velocity as:

w = 2π*f

f = w/2π

f = 7230/6.28

f = 1151 revolutions/s

So, the total revolutions (n) is 12 s was:

n = 12*1151 = 13812 revolutions

3 0

3 years ago