The natural region which holds the Indian subcontinent is jsxoqecxhd ?​

Answers 2

Answer:

it is the Himalayas that holds the Indian subcontinent region

Answer:

The Indian subcontinent, or simply the subcontinent, is a physiographical region in Southern Asia. It is situated on the Indian Plate, projecting southwards into the Indian Ocean from the Himalayas.

Explanation:

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a 1γ=limit ( Σ — −ln a ) a→∞ k=1 k​

Answer:

Using limn11+12+...+1n−ln(n)=γ.

Then

11−12+13−14+...+12n−1−12n=[11+12+13+14+...+12n−1+12n]−2[12+14...+12n]

=[11+12+13+14+...+12n−1+12n]−ln(2n)−[11+12...+1n]+ln(n)+ln2.

Taking the limit we get γ−γ+ln(2).

Step-by-step explanation:

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Using limn11+12+...+1n−ln(n)=γ.

[tex]:\implies[/tex]11−12+13−14+...+12n−1−12n=[11+12+13+14+...+12n−1+12n]−2[12+14...+12n]

[tex]:\implies[/tex][11+12+13+14+...+12n−1+12n]−ln(2n)−[11+12...+1n]+ln(n)+ln2.

Taking the limit we get γ−γ+ln(2).

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From a window (h m ) high above the ground of a house in a street, the angles of elevation and depression of the top and foot of another house on the opposite side of the street are respectively. Show that the height of the opposite house is h(1 + tan a tan ß) m.

Answer:

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Calculate the force which is acting normally on the surface of area 0.3 cm² and pressure exerted on the surface is 40 Nm ².​

Answer:

Solution

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Pressure=

Area

Force

Force=Area×pressure

=3×10

=30N

If the area is made one-third i.e. 1m

2

, then the force would be

Pressure=

Area

Force

Force=Area×pressure

=1×10

=10N

Explanation:

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A potter’s wheel is set into rotation at 100 rpm. It is in the form of a disc of mass10 kg & radius 0.4 m. A lump of clay (to treated as a particle) of mass 1.6 Kg falls & adheres to the wheel at a distance x from its centre. Calculate x if the wheel now rotates at 80 rpm.​

Answer:

Explanation:

Data: f

1

​

=100rpm,f

2

​

=80rpm,M=10kg,R=0.4m,m=1.6kg

I

1

​

=I

wheel

​

=

2

1

​

MR

2

=

2

1

​

(10)(0.4)

2

=0.8kg.m

2

 

The MI of the wheel and the lump of clay is

I

2

​

=I

wheel

​

=mx

2

 

By the principle of conservation of angular momentum,

Iω

1

​

=I

2

​

ω

2

​

 

∴I

1

​

(2πf

1

​

)=I

2

​

(2πf

2

​

)

∴I

2

​

=I

wheel

​

+mx

2

=

f

2

​

f

1

​

​

I

1

​

=

f

2

​

f

1

​

​

I

wheel

​

 

∴mx

2

=(

f

2

​

f

1

​

​

−1)I

wheel

​

=(

80

100

​

−1)(0.8)

=(

4

5

​

−1)(0.8)=0.2kg.m

2

 

v=x

2

=

1.6

0.2

​

=

8

1

​

 

∴x=

8

​

1

​

m=0.3536m

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