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ddd [48]
3 years ago
12

Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei

ghts
Physics
1 answer:
Vikentia [17]3 years ago
6 0

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.
dalvyx [7]

Answer:

(a) W_c=127.008 J

(b) W_g=148.176 J

(c) K.E. = 21.168 J

(d) v=3.4293m.s^{-1}

Explanation:

Given:

  • mass of a block, M = 3.6 kg
  • initial velocity of the block, u=0 m.s^{-1}
  • constant downward acceleration, a_d= \frac{g}{7}

\Rightarrow That a constant upward acceleration of \frac{6g}{7} is applied in the presence of gravity.

∴a=- \frac{6g}{7}

  • height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

F_c= M\times a

F_c=3.6\times (-6)\times\frac{9.8}{7}

F_c=-30.24 N

∴Work by the cord on the block,

W_c= F_c\times d

W_c= -30.24\times 4.2

We take -ve sign because the direction of force and the displacement are opposite to each other.

W_c=-127.008 J

(b)

Force on the block due to gravity:

F_g= M.g

∵the gravity is naturally a constant and we cannot change it

F_g=3.6\times 9.8

F_g=35.28 N

∴Work by the gravity on the block,

W_g=F_g\times d

W_g=35.28\times 4.2

W_g=148.176 J

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

K.E.= W_g+W_c

K.E.=148.176-127.008

K.E. = 21.168 J

(d)

From the equation of motion:

v^2=u^2+2a_d\times d

putting the respective values:

v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }

v=3.4293m.s^{-1} is the speed when the block has fallen 4.2 meters.

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3 years ago
Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3
RideAnS [48]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 12.6 N

d = 15.4 m

\theta = 15.3 degree

so we will have

W = 12.6 \times 15.4 cos15.3

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A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?
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Answer:

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==> If the box is at rest on the table, then it is not accelerating.

==> Since it is not accelerating, I can say that the forces on it are balanced.

==> That means that the sum of all forces acting on the box is zero,
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==> This in turn means that all of the horizontal forces are balanced,
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Horizontal forces:
sliding friction, somebody pushing the box

All of the forces on this list must add up to zero. So ...

(sliding friction force) = (pushing force), in the opposite direction.

If nobody pushing the box, then sliding friction force = zero.

Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)

All of the forces on this list must add up to zero, so ...

(Gravitational force down) + (normal force up) = zero

(Gravitational force down) = -(normal force up) .
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