The area of the curved surface area of a cylinder is three times the base area and height of the cylinder exceeds the radius by 7cm .Find the volume of the cylinder

Answer:

भारताचा पूर्व पश्चिम, उत्तर दक्षिण विस्तार ७,३०० आहे.

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• दुसर् यापासून पूर्वेकडे पसरलेल्या, भारत 68°7'ई आणि 97° 25'ई रेखावृत्तांच्या दरम्यान विस्तारलेला आहे.
• अरुणाचल प्रदेशापासून कच्छपर्यंत पूर्व-पश्चिम भाग सुमारे २,९०० कि.मी. भारताचा रेखावृत्तीय विस्तार पश्चिमेकडे ६८°७°ई.
• पर्यंत असून पूर्वेकडे ९७°२५° ई इतका आहे.
• 82°30° ई रेखावृत्त भारतीय मानक मेरिडियन म्हणून घेतले जाते.
• भारताचे एकूण क्षेत्रफळ ३२,८७२६३ चौ.कि.मी.
• भारताच्या मुख्य भूमीचे पूर्व पश्चिम कमाल अंतर ३,००० कि.मी. आहे.
• पोर्तुगीज साम्राज्यामुळे पोर्तुगीज साम्राज्याने पोर्तुगीज साम्राज्याला ग्लोबचा बराचसा भाग शोधून काढला आणि त्याचा नकाशा तयार केला आणि केप ऑफ गुड होपमार्गे भारताकडे येणारा सागरी मार्ग शोधण्याचा उल्लेखनीय प्रवास,
• जसे की पूर्व व पश्चिमेपर्यंत सागरी मार्ग शोधला.

Answer:

answer is in attachment

Answer:

There are two types of water sampling strategies-

1. Discrete samples.
2. Composite samples.

Discrete sample, also known as grab sample, is a single sample collected in an individual container. The sample is representative of the chemistry only at the time and place at which the sample was taken.

Composite sample consists of a series of smaller samples collected at a predetermined time or after predetermined flow and mixed in the same container.

Answer:

vertical component of velocity is zero at highest point

so, v= ucos30

v=15× 3½/2

v= 7.5 underoot 3

Given:

Initial speed of the arrow [tex]=15m/s[/tex]

Elevation with respect to horizontal [tex]=30^{o}[/tex]

To find:

Speed at the highest point of the trajectory.

Solution:

Step 1

For a projectile thrown with an initial velocity [tex]u[/tex] at an angle [tex]\alpha[/tex] with respect to the ground, the maximum height reached by the projectile is given by

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

The initial velocity of projection can be resolved into its rectangular components namely [tex]u.cos\alpha[/tex] horizontally and [tex]u.sin\alpha[/tex] vertically.

If we take a look at the vertical component of velocity, this component is responsible for the height achieved by the object.

Just like a ball, which when thrown up, with some initial velocity gets acted upon by the acceleration due to gravity in the downward direction and at the maximum height that is when the potential energy becomes maximum, The Final velocity of the object becomes zero [tex]0m/s[/tex].

Hence,

The value of vertical component of the velocity at the highest point of the projectile will be

[tex]V_{v} =0m/s[/tex]

Step 2

Now,

If we consider the horizontal component of the velocity, [tex]u.cos\alpha[/tex], this component of velocity is responsible for making the object travel the maximum range which is given by

[tex]R=\frac{u^{2}.sin2\alpha }{g}[/tex]

In the horizontal motion of the object, we know that since no acceleration acts on the body during the flight hence, velocity at the beginning of the projectile is exactly equal to the velocity of the projectile at the end of the flight.

Hence, the horizontal component of velocity of the object at the highest point of the projectile is

[tex]V_{h}=u.cos\alpha[/tex]

[tex]V_{h} =15.cos(30^{o} )[/tex]

[tex]V_{h}=15(\frac{\sqrt{3} }{2} )[/tex]

[tex]V_{h}=12.99m/s[/tex]

Step 3

Hence, the velocity of the projectile at the highest point will be

[tex]V=\sqrt{V_{h} ^{2} +V_{v} ^{2} }[/tex]

[tex]V=\sqrt{(12.99)^{2} +(0)^{2} }[/tex]

[tex]V=\sqrt{(12.99)^{2} }[/tex]

[tex]V=12.99m/s[/tex]

Final answer:

Hence, the velocity of the projectile at the highest point of the projectile is 13 m/s (approx).

Answer:

a = 5.61

b = 12.46

c = 45.87

d = 61.16

e = 69.76

f = 141.5

g = 704.92

h = 38.361