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evablogger [386]
3 years ago
5

A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen witho

ut fracture, given that the applied load is 750 N (169 lbf), the flexural strength is 105 MPa (15,000 psi), and the separation between load points is 75.0 mm (2.95 in.).
Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

The answer is 4.7*10^{-3} m

Explanation:

The equation for maximum ratio is given

r max = \sqrt[3]{\frac{F*L}{ro*pi} } = \sqrt[3]{\frac{750 (45*10^{-3})}{(105 * 10^{6})*pi }  }

rmax = 4.7*10^{-3} m

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A 5 kg ball is suspended on one cable. Calculate the tension in the cable
ludmilkaskok [199]

Answer:

49N

Explanation:

F=ma

We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2

F=5kg*9.8m/s^2

 = 49N

Hope this helps!

7 0
2 years ago
Hello people ~
Novosadov [1.4K]

Answer:

Copper

Explanation:

Capacitance is directly proportional to dielectric constant

Aluminium and zinc are highly reactive and have high dielectric contact.

Copper has less dielectric constant hence capacitance will decrease

3 0
2 years ago
Two or more velocities add by ____?<br><br> Plz help
VladimirAG [237]
By vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities). 
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
R= \sqrt{(R_x)^2+(R_y)^2}
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
\tan \alpha =  \frac{R_y}{R_x}
where \alpha is its direction with respect to the x-axis.
3 0
3 years ago
Read 2 more answers
Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges becom
jeka57 [31]

Answer:

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

       

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

3 0
3 years ago
Two particles, one with charge -5.45 × 10^-6 C and one with charge 4.39 × 10^-6 C, are 0.0209 meters apart. What is the magnitud
Levart [38]

Explanation:

Given that,

Charge 1, q_1=-5.45\times 10^{-6}\ C

Charge 2, q_2=4.39\times 10^{-6}\ C

Distance between charges, r = 0.0209 m

1. The electric force is given by :

F=k\dfrac{q_1q_2}{r^2}

F=9\times 10^9\times \dfrac{-5.45\times 10^{-6}\times 4.39\times 10^{-6}}{(0.0209)^2}

F = -492.95 N

2. Distance between two identical charges, r=0.0209\ m

Electric force is given by :

F=\dfrac{kq_3^2}{r^2}

q_3=\sqrt{\dfrac{Fr^2}{k}}

q_3=\sqrt{\dfrac{492.95\times (0.0209)^2}{9\times 10^9}}

q_3=4.89\times 10^{-6}\ C

Hence, this is the required solution.

3 0
2 years ago
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