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kifflom [539]
3 years ago
9

You can divide a surface by drawing a line through it

Engineering
2 answers:
tatuchka [14]3 years ago
8 0

Answer:

yes you can

Explanation:

hope this helps, have a good day :-)

SpyIntel [72]3 years ago
6 0

Answer:

T

Explanation:

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I WILL GIVE 20 POINTS!!
Alex777 [14]

Answer:

Use a resume header

Explanation:

Create a Summary

Research industry, employer keywords

there are some hints okay

5 0
3 years ago
Read 2 more answers
What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF
exis [7]

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

           C=capacitance

           V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C=\frac{Q}{V}

=500×10^{-12}×\frac{1}{250}

=2×10^{-12}

=2 pF

3 0
3 years ago
What happens when the arms of the milky move away from the center of the galaxy
Alina [70]
Well this question is though because we have never seen such a thing ! and to be quite frank when that happens , nothing good comes from it. Black Holes
6 0
3 years ago
A material has the following properties: Sut = 275 MPa and n = 0.40. Calculate its strength coefficient, K.
Tems11 [23]

Answer:

The strength coefficient is K = 591.87 MPa

Explanation:

We can calculate the strength coefficient using the equation that relates the tensile strength with the strain hardening index given by

S_{ut}=K \left(\cfrac ne \right)^n

where Sut is the tensile strength, K is the strength coefficient we need to find and n is the strain hardening index.

Solving for strength coefficient

From the strain hardening equation we can solve for K

K = \cfrac{S_{ut}}{\left(\cfrac ne \right)^n}

And we can replace values

K = \cfrac{275}{\left(\cfrac {0.4}e \right)^{0.4}}\\K=591.87

Thus we get that the strength coefficient is K = 591.87 MPa

6 0
3 years ago
The sliders A and B are connected by a light rigid bar of length l = 20 in. and move with negligible friction in the slots, both
DedPeter [7]

Answer:

Explanation:

Given:

- The Length of the rigid bar L = 20 in

- The position of slider a, x_a = 16 in

- The position of slider b, y_b

- The velocity of slider a, v_a = 3 ft /s

- The velocity of slider b, v_b

- The acceleration of slider a, a_a

- The acceleration of slider b, a_b

Find:

-Determine the acceleration of each slider and the force in the bar at this instant.

Solution:

- The relationship between the length L of the rod and the positions x_a and x_b of sliders A & B is as follows:

                               L^2 = x_a^2 + y_b^2   ....... 1

                               y_b = sqrt( 20^2 - 16^2 )

                               y_b = 12

- The velocity expression can derived by taking a derivation of Eq 1 with respect to time t:

                               0 = 2*x_a*v_a + 2*y_b*v_b

                               0 = x_a*v_a + y_b*v_b   ..... 2

                               0 = 16*36 + 12*v_b

                               v_b = - 48 in /s = -4 ft/s

- Similarly, the acceleration expression can be derived by taking a derivative of Eq 2 with respect to time t:

                               0 = v_a^2 + x_a*a_a + v_b^2 + y_b*a_b

                               0 = 9 + 4*a_a/3 + 16 + a_b

                               4*a_a/3 + a_b = -25

                               4*a_a + 3*a_b = -75  .... 3

- Use dynamics on each slider. For Slider A, Apply Newton's second law of motion in x direction:

                               F_x = m_a*a_a

                               P - R_r*16/20 = m_a*a_a

                               

- For Slider B, Apply Newton's second law of motion in y direction:

                               F_y = m_b*a_b

                               - R_r*12/20 = m_b*a_b

- Combine the two dynamic equations:

                               P - 4*m_b*a_b / 3 = m_a*a_a

                               3P = 3*m_a*a_a + 4*m_b*a_b  ... 4

- Where,                  P = Is the force acting on slider A

                               P , m_a and m_b are known quantities but not given in question. We are to solve Eq 3 and Eq 4 simultaneously for a_a and a_b.                    

5 0
3 years ago
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