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ludmilkaskok [199]
2 years ago
12

A rectangular block has the density of 350g/cm3 the dimensions are 3.5 6.cm 2.5

Physics
1 answer:
LekaFEV [45]2 years ago
7 0

Answer:

18375g

Explanation:

\boxed{density =  \frac{mass}{volume} }

∴ mass = density \times volume

Let's find the volume of the rectangular block.

Volume

= length ×breadth ×height

= 3.5 ×6 ×2.5

= 52.5cm³

Mass of the block

= 350(52.5)

= 18375g

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1. What is the potential energy of a 5.0-kg
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Answer:

  C.  98 J

Explanation:

The appropriate formula is ...

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5 0
3 years ago
A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.
matrenka [14]

Answer:

Part a)

T = 42 N

Part b)

v_f = 11.8 m/s

Part c)

t = 1.7 s

Part d)

F = 159.7 N

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

mg - T = ma

for uniform cylinder we will have

TR = I\alpha

so we have

T = \frac{1}{2}MR^2(\frac{a}{R^2})

T = \frac{1}{2}Ma

now we have

mg = (\frac{M}{2} + m)a

a = \frac{mg}{(\frac{M}{2} + m)}

a = \frac{15 \times 9.81}{(6 + 15)}

a = 7 m/s^2

now we have

T = \frac{12 \times 7}{2}

T = 42 N

Part b)

speed of the bucket can be found using kinematics

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(7)(10)

v_f = 11.8 m/s

Part c)

now in order to find the time of fall we can use another equation

v_f - v_i = at

11.8 - 0 = 7 t

t = 1.7 s

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

F = T + Mg

F = 42 + (12 \times 9.81)

F = 159.7 N

5 0
3 years ago
Which data set has the largest range?
blsea [12.9K]

Answer:

Line the numbers from smallest to largest the subtract the smallest from the largest numbers.

3 0
3 years ago
A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels
katovenus [111]

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v = \sqrt{\frac{T}{\mu}}

Here,

v = Velocity

\mu= Linear density (Mass per  unit length)

T = Tension

Rearranging to find the Period we have that

T = v^2 \mu

T = v^2 (\frac{m}{L})

As we know that speed is equivalent to displacement in a unit of time, we will have to

T = (\frac{L}{t}) ^2(\frac{m}{L})

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T = 5.54N

Therefore the tension is 5.54N

8 0
3 years ago
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