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3 years ago

Do not answer pls thank you

Engineering

2 answers:

erma4kov [3.2K]3 years ago

6 0

Answer:

it is making me write something

Explanation:

hope ur having a good night

astra-53 [7]3 years ago

4 0

The answer is answered! Explanation:

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A company intends to market a new product and it estimates that there is a 20% chance that it will be first in the market

Stells [14]

The EMV - Ending Market Value is given as:
$2,400,000.

<h3>How is the EMV Arrived At?</h3>

The EMV is given as:

BMV x (i + r); Where

BMV is the Beginning Market Value; and

r is the interest rateor percentage given.

Hence the EMV = 2,000,000 x ( 1 + 20%)

= 2,000,000 x 1.2

= $ 2, 400,000.

It is to be noted that the BMV is the Beginning Market Value which is the value of an investment at the start of the business period.

Learn more about Market Value at:

brainly.com/question/1350233

4 0

2 years ago

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

8 0

3 years ago

As described in "A Note About Bacterial Reproduction -- and the "Culture Bias,"" the organism Epulopisciumdoes not divide by bin

zloy xaker [14]

Answer:

Explanation:

The best method that will yield significantly more accurate result is to use spectrophotometer to read the turbidity of the sample and increase in turbidity is associated with increase biomass.

5 0

3 years ago

What type of intersection is this?

mote1985 [20]

Diverging Diamond Interchange

6 0

3 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$