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3 years ago

12

Which of these gets so bright as to be seen in daylight at times? mars saturn jupiter mercury venus?

2 answers:

fenix001 [56]3 years ago

8 0

While these could all be seen with a telescope during daytime, Venus<span> is the only one that could be viewed with the naked eye in daylight.</span>

tamaranim1 [39]3 years ago

5 0

Venus sometimes does that.

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Which statement is true? Which statement is true? An orbital that penetrates into the region occupied by core electrons is more

masya89 [10]

Hey there!:

Here the Statement - D is correct.

Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate. Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.

Hope this helps!

6 0

3 years ago

A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60

german

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

$m_1$ = Mass of glider without person = 680-60 kg

$v_1$ = Gliders speed just after the skydiver lets go

$m_2$ = Mass of person = 60 kg

$v_2$ = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

$m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s$

The gliders speed just after the skydiver lets go is 34 m/s

3 0

3 years ago

Which disease do you think is most easily spread? 5 Answers

Anika [276]

Answer:

coronavirus

sida

tuberculosis

sifilis

epatitis

Explanation:

7 0

3 years ago

A locomotive has a mass of 200,000kg. It is moving at 4.5m/s. Find its momentum .

anastassius [24]

Momentum (p) = mass × velocity

P= 200,000×4.5

P= 900,000 .... answer !!

4 0

3 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