Which among the following has positive value of electron gain enthalpy? a) Se b) Br c) Hed) Na​

• We know a relation

• R tanθ = 4 H

Given: R = 2H ; u = 20m/s

• That implies, H/R = tanθ/4

⇒ H/2H = tanθ/4

• 1/2 = tanθ /4

We'll get tanθ= 2

Now we'll draw a triangle to calculate all trignometric frnctions easily

You can see the triangle in the Attachment.

From the triangle sinθ = [tex]\frac{2}{\sqrt[]{5} }[/tex] ; cosθ = [tex]\frac{1}{\sqrt[]{5} }[/tex]

H = [tex]\frac{u^2sin^2theta}{2g}[/tex]

H = [tex]\frac{400 * (\frac{2}{\sqrt{5} })^2 }{20} = \frac{4}{5} *20 = 16m[/tex]

⇒Anagh :}

Given:

Initial velocity [tex]=20m/s[/tex]

Acceleration due to gravity [tex]=10m/s^{2}[/tex]

To find:

Maximum height reached by the projectile.

Solution:

Step 1

For a projectile released from a point with some initial velocity as [tex]u[/tex] and the angle made by the projectile with the horizontal,

The height reached by the projectile [tex]H=\frac{u^{2} .sin^{2}\alpha }{2g}[/tex]  

The range of the projectile is [tex]R=\frac{u^{2}.sin 2\alpha }{g}[/tex]  

Step 2

Now,

According to the question,

The height reached by the projectile is equal to half the range of the projectile. Hence, we get

[tex]\frac{u^{2} .sin^{2}\alpha }{2g}=\frac{1}{2}\frac{u^{2}.sin 2\alpha }{g}[/tex]

[tex]sin^{2} \alpha =sin2\alpha[/tex]

[tex]sin^{2} \alpha =2.sin\alpha .cos\alpha[/tex]

[tex]sin\alpha =2.cos\alpha[/tex]

[tex]tan\alpha =2[/tex]

We know,

[tex]sec^{2} \alpha =1+tan^{2} \alpha[/tex]

[tex]sec^{2} \alpha =5[/tex]

Hence,

[tex]cos^{2} \alpha =\frac{1}{5}[/tex]

We know,

[tex]sin^{2} \alpha =1-cos^{2} \alpha[/tex]

[tex]sin^{2} \alpha =\frac{4}{5}[/tex]

Step 3

Now,

We have, [tex]u=20m/s[/tex]   [tex];[/tex] [tex]sin^{2} \alpha =\frac{4}{5}[/tex]    [tex];[/tex] [tex]g=10m/s^{2}[/tex]

We get, the height reached by the projectile as

[tex]H=\frac{u^{2} .sin^{2}\alpha }{2g}[/tex]

[tex]H=\frac{(20)^{2}(\frac{4}{5} ) }{2(10)}[/tex]

[tex]H=16m[/tex]

Final answer:

Hence, the maximum height reached by the projectile is 16 meter.

Answer:

The tone is an angry one.

Explanation:

In the question, we have been asked to comment on the tone of the speaker when she says, ‘Will you please look at me when I’m speaking to you, Amanda!​

From the sentence, we can figure out that the speaker is trying to convey something to Amanda and she is not paying attention. So the speaker is getting angry and asking her to look at the speaker while she is speaking to Amanda.

Answer:

the tone is anger

Explanation:

because the person named Amanda is ignoring him/her

Answer:

pi orbital

Explanation:

bec there is weak pi bond that's why comparative sigma bond

Explanation:

• [tex]B2+[/tex]  has 10 electrons  

• Its electronic configuration is [tex]sigma (1s), 2 sigma*(1s) ,2 sigma(2s), 2 sigma* (2s) ,2 pi (2p) x1, Pi 2p y1[/tex]

• Here unpaired electrons are found and due to the presence of unpaired electrons, the nature is paramagnetic.

• Here the highest occupied molecular orbital is of [tex]\pi[/tex] orbital.
• pi bonds are covalent chemical bonds.
• They can form a double and triple bonds but do not form single bonds.
• They involve complete overlapping of two lobes of atomic orbital with the other.

Answer:

cotton is exposed to the air or black carbon it turns into black

• The fine cotton thread-like structure is a member of kingdom fungi and its name is Rhizopus stolonifer.
• The branching fine cotton thread is called hyphae and all the hyphae are together known as mycelium.
• The bread mould forms spores inside a sporangium. When it breaks open, they germinate to form a new mould.
• The new mould appears black in colour.