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katovenus [111]
3 years ago
8

6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

12.2 N.Find the speed of the block after it has at 2.5m
Physics
1 answer:
neonofarm [45]3 years ago
4 0

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed  = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s

\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5

 3.3 v² = 30.5

     v² = 9.242

      v = 3.04 m/s

speed of the block is equal to 3.04 m/s

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3 years ago
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A diverging lens with a focal length of 14 cm is placed 12 cm to the right of a converging lens with a focal length of 21 cm. An
IRINA_888 [86]

Answer:

-0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

0.023 m  right of diverging lens

Explanation:

given data

focal length f2 = 14 cm = -0.14 m

Separation s = 12 cm = 0.12 m

focal length f1 = 21 cm = 0.21 m

distance u1 = 38 cm

to find out

final image be located and Where will the image

solution

we find find  image location i.e v2

so by lens formula v1 is

1/f = 1/u + 1/v     ...............1

v1 = 1/(1/f1 - 1/u1)

v1 = 1/( 1/0.21 - 1/0.38)

v1 = 0.47 m

and

u2 = s - v1

u2 = 0.12 - 0.47

u2 = -0.35

so from equation 1

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.35)

v2 = -0.233 m

so -0.233 m left of diverging lens and ( 0.12 - 0.233 ) = -113 m left of conversing

and

for Separation s = 45 cm = 0.45 m

v1 = 1/(1/f1 - 1/u1)

v1 =0.47 m

and

u2 = s - v1

u2 = 0.45 - 0.47 =- 0.02 m

so

v2 = 1/(1/f2 - 1/u2)

v2 = 1/(-1/0.14 + 1/0.02)

v2 = 0.023

so here 0.023 m  right of diverging lens

6 0
3 years ago
A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t
viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

mv+MV=(m+M)v'

Here , v' is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s

Hence , this is the required solution .

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3 years ago
how much work did the movers do (horizontally) pushing a 41.0- kg crate 10.3 m across a rough floor without acceleration, if the
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W=2485.65 J ,work did the movers do friction force a 41.0- kg crate 10.3 m across a rough floor without acceleration

<h3>What is a basic friction force?</h3>

Two surfaces that come into contact and slide against one another produce a force known as frictional force. Several aspects that influence the frictional force include: The surface texture as well as the amount of force attracting them together have the most effects on these forces.

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It generates heat, which is useful for warming our bodies or specific areas of any object. Power is also lost as a result. It makes noise during every operation. We are able to walk, run, dance, etc. due to friction.

To know  more about friction force visit:

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