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2 years ago

9

Astronauts wear liquid cooled space suits to keep their body temperature moderate. One

1 answer:

Anton [14]2 years ago

6 0

Answer:

D. Because they are using space technology on a shirt so people can wear it on earth as well

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An imaginary line perpendicular to a reflecting surface is called _________.

n200080 [17]

<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>

5 0

2 years ago

Read 2 more answers

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

3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of

Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

F is applied force
Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0

1 year ago

I WILL GIVE BRAILYEST!!! What is the mass of an object moving at a velocity of 5 m/s if the momentum of the object is 50 kg•m/s?

lianna [129]

Answer:

a. 250kg I think it's the right answer. hope it helps:)

7 0

2 years ago

Read 2 more answers

A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

2 years ago