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2 years ago

9

What chemical phenomenon accounts for the elasticity seen in solids?

1 answer:

Natali [406]2 years ago

3 0

Answer:

Hydrogen bonds is answer

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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

<u>Given :</u>

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

<u>Let's Slove :</u><u> </u>

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

3 0

1 year ago

I need help with #24 ASAP .. I have to get it done today.. I need help with #24 right now

Savatey [412]

Answer:

Explanation: the temperature range is at 1,400

8 0

1 year ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Bond [772]

The y-component of the acceleration is $0.33 m/s^2$

Explanation:

The y-component of the acceleration is given by:

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

$u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s$

where

u = 2.25 m/s is the initial velocity

$\theta_1 = 50.0^{\circ}$ is the initial direction

$v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s$ , where

v = 4.65 m/s is the final velocity

$\theta_2 = 120.0^{\circ}$ is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

$a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

2 years ago

To protect a material from the influence of an external magnetic field, the material should be kept in?

Evgen [1.6K]

To protect a material from the influence of an external magnetic field, the material should be kept in soft iron ring.

So the correct answer is A.

Hope this helps,

Davinia.

5 0

2 years ago