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3 years ago

6

You are discussing a material that has a strong resistance to the flow of electrons. This is a description of a(n)

2 answers:

cupoosta [38]3 years ago

5 0

C. Insulator

It COULD be semi-insulator but i'm sure its C

eimsori [14]3 years ago

4 0

Insulator. the right answer'

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The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

BartSMP [9]

Answer:

hello your question is incomplete attached below is the missing part

answer : short period oscillations frequency = 0.063 rad / sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= $( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0$

where :

$w_{np} = Natural frequency of plugiod oscillation$

$\alpha _{p} = damping ratio of plugiod oscilations$

comparing the general form with the given equation

$w^{2} _{np}$ = 18.2329

$w^{2} _{ns} = 0.003969$

hence the short period oscillation frequency ( $w_{ns}$ ) = 0.063 rad/sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

8 0

3 years ago

How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?

Marrrta [24]

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by $V = R I$

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

$115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes$

8 0

3 years ago

Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.

-The equations of motion from Newton's Second Law are:

$mBg-T=mBa$ where a is the acceleration.

#Substituting for $T$ (tension) gives:

$mBg-mAsin\theta-mAa=mBa$

#and solving for $a:$

$a=\frac{mBg-mAgSin\theta}{mA+mB}$ which is the system's acceleration.

8 0

4 years ago

Martha wants to calculate an object's velocity. What will she need to do?

rjkz [21]

Answer:

you divide the distance by the time it takes to travel that same distance, then you add your direction to it.

4 0

2 years ago

A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all

timurjin [86]

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, $a=5\times 10^{-4}\ m^2$

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}$

$P=1.4\times 10^5\ Pa$

We know that the atmospheric pressure is given by :

$P_o=1.013\times 10^5\ Pa$

So, the pressure is given by :

$p=P+P_o$

$p=1.4\times 10^5+1.013\times 10^5$

$p=2.41\times 10^5\ Pa$

Hence, this is the required solution.

7 0

3 years ago