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3 years ago

Who were the sepoys?

Physics

2 answers:

Nat2105 [25]3 years ago

7 0

Answer:

A. Indians who were part of the East India Company’s private army.

Explanation:

The sepoys were Indian soldiers who were recruited into the Company's army. Just before the rebellion, there were over 300,000 sepoys in the army, compared to about 50,000 British. The forces were divided into three presidency armies: Bombay, Madras, and Bengal.

Sergeeva-Olga [200]3 years ago

6 0

Answer:

A-Indians who were part of the East India Company’s private army

Explanation:

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A mass of 2 kg is attached to a spring and placed on a horizontal surface. The spring has a spring constant of 20 N/m, and the s

Sergio [31]

The speed of the mass : <em><u>v = 0.316 m/s</u></em>

<h3>Further explanation</h3>

The energy used to press a spring is included as the potential energy

Can be formulated:

$\displaystyle E_p=\frac{1}{2}kx^2$

Ep= potential energy

k = spring constant

x = change in spring length

If the spring is released from its pressure, this potential energy will turn into kinetic energy

so applies the law of conservation of mechanical energy (Em)

Ek = Ep

A mass of 2 kg is attached to a spring, a spring constant of 20 N/m, and the spring is compressed 0.1 m past its natural length.

m = 2 kg

k = 20 N/m

x = 0.1 m

$\displaystyle Ep=Ek\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\kx^2=mv^2\\\\20\times0.1^2=2\times v^2\\\\v^2=0.1\\\\v=\sqrt{0.1}\\\\v=0.316~m/s$

<h3>Learn more</h3>

Hooke's law

brainly.com/question/2648431

Keywords : spring,mass, spring constant,compressed position

5 0

3 years ago

What type of physical weathering is most common in a river

tia_tia [17]

Answer:

Exfoliation or Unloading. As upper rock portions erode, underlying rocks expand. ...

Thermal Expansion. Repeated heating and cooling of some rock types can cause rocks to stress and break, resulting in weathering and erosion. ...

Organic Activity. ...

Frost Wedging. ...

Crystal Growth.

Explanation:

3 0

4 years ago

Read 2 more answers

The greatest pull of a magnet is near its poles.<br><br> True<br> False

DIA [1.3K]

False because opposites attract. :)

6 0

3 years ago

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Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

3 years ago

Can uh help in in this question step by step

Luda [366]

Initial velocity=u=72km/h

Convert to m/s

$\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s$

Final velocity=v=0m/s
Time=2s=t

$\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}$

$\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}$

$\\ \sf \longmapsto Acceleration=a=-10m/s^2$

Distance be s

Using second equation of kinematics

$\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2$

$\\ \sf \longmapsto s=40+(-20)$

$\\ \sf \longmapsto s=40-20$

$\\ \sf \longmapsto s=20m$

Now

Mass=m=5000kg

Using newtons second law

$\\ \sf \longmapsto Force=ma$

$\\ \sf \longmapsto Force=5000(-10)$

$\\ \sf \longmapsto Force=-50000N$

Force is in opposite direction so its negative

$\\ \sf \longmapsto Force=50kN$

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3 years ago

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