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1 year ago

Hey answer quick for 20 points

Engineering

2 answers:

SSSSS [86.1K]1 year ago

4 0

Thank you!!!!!!!!!!!!!!!!

creativ13 [48]1 year ago

3 0

Answer:

<h2>Hey.....Thanks but FYI this is only 8 points....If u want to put 20....try putting 40...</h2>

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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

2 years ago

Please help me fast, I don’t have time

Anna71 [15]

Answer: precision

Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer

5 0

2 years ago

Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mec

olga55 [171]

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

Explanation:

Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg

6 0

2 years ago

Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

7 0

2 years ago

You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you

telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l = 35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L = 0.00841549 / 1.4

L = 6 × 10⁻³ H

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV

Therefore, the emf generated in the coil is 18 mV

7 0

2 years ago