which personality type most enjoys accomplishment and competition

Here is the answer.

I hope it will help you.

Answer:

protection of the human in the society

Answer:

भावार्थ : कवि रामधारीसिंह दिनकर कहते हैं कि यह मनुष्य सृष्टि का श्रृंगार, ज्ञान और विज्ञान तथा आलोक का आगार है । आकाश से पाताल तक की सब-कुछ जानकारी इसे है । परंतु यह उसका सही परिचय नहीं है और न ही उसका श्रेय है ।

• 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