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3 years ago

If an older multimeter is set to display its reading in megohms, what must you do to determine the correct value? Please help.

Engineering

1 answer:

Gekata [30.6K]3 years ago

8 0

Answer:

idk

Explanation:

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Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi

lawyer [7]

Answer:

$\vec F_{A} = -67500\,N\cdot (i + j)$

Explanation:

The position of each point are the following:

$A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)$

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

$\vec F_{A} = \vec F_{AB} + \vec F_{AC}$

$\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j$

$\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)$

$\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)$

$\vec F_{A} = -67500\,N\cdot (i + j)$

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3 years ago

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3 years ago

Describe how you could engineer the situation to produce even more friction and heat

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2 years ago

How do you build a engine best anwser will get marked as brainlist

marysya [2.9K]

Answer:

It is best to soak the new lifters in engine oil overnight in order to build a car engine from scratch, regardless of its length. Ensure all rings of the pistons are installed correctly. You should flip the block over and carefully install the top half of the main rod and bearing parts. Get the block out of your way and flip it over again. By flipping over the block, you’ll see the full effect. Ensure that the lifters are installed correctly.

Explanation:

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2 years ago

The velocity field of a flow is given by where x and y are in feet. Determine the fluid speed at points along the x axis; along

lora16 [44]

Answer:

Using the formula

V =20y/(x^2+y^2)^1/2 - 20x/(x^2+y^2)^1/2

Hence fluid speed at x axis =20x/(x^2+y^2)^1/2

While the fluid speed at y axis =20y/(x^2+y^2)^1/2

Now the angle at 1, 5

We substitute into the formula above

V= 20×5/(1+25)^1/2= 19.61

For x we have

V = 20× 1/(1+25)^1/2= 3.92

Angle = 19.61/3.92= 5.0degrees

Angel at 5, and 2

We substitute still

V = 20×5/(2+25)^1/2=19.24

At 2 we get

V= 20×2/(2+25)^1/2=7.69

Dividing we get 19.24/7.69=2.5degrees

At 1 and 0

V = 20/(1)^1/2=20

At 0, v =0

Angel at 2 and 0 = 20degrees

At 5 and 2

V = 100/(25+ 4)^1/2=18.56

At x = 2

40/(√29)=7.43

Angle =18.56/7.43 = 2.49degrees.

6 0

3 years ago

Read 2 more answers