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timurjin [86]
2 years ago
11

A 2 kg ice cube is released from point A and slides on a frictionless track as shown in the figure. Determine the speed of the c

ube at points B and C.

Physics
1 answer:
kherson [118]2 years ago
8 0

According to the conservation of energy

  • Potential energy at any given instance is equal to the Kinetic energy as energy can neither be created nor be destroyed

Mass is 2kg=m

#A

h=5m

  • PE=mgh

PE

  • 2(9.8)(5)
  • 10(9.8)
  • 98J

Now

  • KE=98J
  • 1/2mv²=98J
  • 1/2×2v²=98J
  • v²=98J
  • v=√98
  • v=9.4m/s

#B

h=3.2m

PE:-

  • 2(3.2)(9.8)
  • 6.4(9.8)
  • 62.7J

Now

  • KE=62.7J
  • 1/2mv²=62.7
  • v²=62.7
  • v=√62.7
  • v=7.9m/s

#C

h=2m

PE

  • 2(2)(9.8)
  • 4(9.8)
  • 39.2J

Now

  • v²=39.2
  • v=√39.2
  • v=6.3m/s
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bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

So with y(1) = v - 4.9 = 0 we have  

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

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3 years ago
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For a certain RLC circuit the maximum generator EMF is 125 V and the maximum current is 3.20 A. If le a) the impedance of the ci
MAXImum [283]

Answer:

Part (i)

Z = 39.06 ohm

Part (ii)

R = 21.7 ohm

Explanation:

a) here we know that

maximum value of EMF = 125 V

maximum value of current = 3.20 A

now by ohm's law we can find the impedence as

z = \frac{V_o}{i_o}

now we will have

z = \frac{125}{3.20} = 39.06 ohm

Part b)

Now we also know that

\frac{R}{z} = cos\theta

\theta = 0.982 rad = 56.3 degree

now we have

\frac{R}{39.06} = cos56.3

R = 21.7 ohm

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3 years ago
A bag of sand has a density of 45 g/cm3 and a mass of 15 kg. How much space does the sand take up?
Aleks [24]

Volume = mass/density

 

Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>

3 0
3 years ago
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HELP ASAP!!
andriy [413]

Answer:

This is false becuase different object weigh different

Thank you!

Explanation:

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3 years ago
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