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1 year ago

Question is written in following attached screenshot.

Engineering

1 answer:

Juli2301 [7.4K]1 year ago

7 0

Answer:

jjj jk to the day of the day of the day of the day of the day

Explanation:

hey hi I am a br of August and I am not able to join the meeting and will

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Consider a two-dimensional incompressible velocity potential phi = ???????? cos theta + ????????theta, where B and L are constan

scoundrel [369]

solution:

Increasing the magnification and decreasing the field view

we are given:

a(t)=4*t^2-)

where t= time in seconds, and a(t) = acceleration as a function of time.

and

x(0)=-2 )

x(2) = -20 )

where x(t) = distance travelled as a function of time.

need to find x(4).

from (1), we express x(t) by integrating, twice.

velocity = v(t) = integral of (1) with respect to t

v(t) = 4t^3/3 - 2t + k1 )

where k1 is a constant, to be determined.

integrate (4) to find the displacement x(t) = integral of (4).

x(t) = integral of v(t) with respect to t

= (t^4)/3 - t^2 + (k1)t + k2 ) where k2 is another constant to be determined.

from (2) and (3)

we set up a system of two equations, with k1 and k2 as unknowns.

x(0) = 0 - 0 + 0 + k2 = -2 => k2 = 2 )

substitute (6) in (3)

x(2) = (2^4)/3 - (2^2) + k1(2) -2 = -20

16/3 -4 + 2k1 -2 = -20

2k1 = -20-16/3 +4 +2 = -58/3

k1 = -29/3 )

thus substituting (6) and (7) in (5), we get

x(t) = (t^4)/3 - t^2 - 29t/3 + 2 )

which, by putting t=4 in (8)

x(4) = (4^4)/3 - (4^2 - 29*4/3 +2

= 86/3, or

= 28 2/3, or

= 28.67 (to two places of decimal)

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2 years ago

1. In a series circuit the sum of all voltage drops produced by the loads in the circuit

GarryVolchara [31]

Answer:

ik the answer but I'm busy rn

5 0

2 years ago

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19

Pepsi [2]

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

$V = A*t$

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

$V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}$

Now, using the density we can find the mass:

$m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g$

Finally, with the Avogadros number ( $N_{A}$ ) and with the atomic mass (A) we can find the number of atoms (N):

$N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms$

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

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valentinak56 [21]

Carbonation is more of a healer to the engine

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