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Morgarella [4.7K]
3 years ago
11

What is the acceleration of a cabinet of mass 45 kilograms if Jake and Ted push it by applying horizontal force of 25 newtons an

d 18 newtons respectively in the same direction
Physics
1 answer:
sladkih [1.3K]3 years ago
3 0

Answer:

a=0.96\ m/s^2

Explanation:

Given that,

Mass of cabinet, m = 45 kg

Two horizontal force of 25 newtons and 18 newtons respectively in the same direction.

When the forces are acting in same direction, the net force is equal to the sum of forces i.e.

F = 25 N + 18 N = 43 N

Let a is the aceleration of the cabinet

So,

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{43}{45}\\\\a=0.96\ m/s^2

So, the acceleration of the cabinet is 0.96\ m/s^2.

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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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3 years ago
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A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti
melisa1 [442]
1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
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4 0
3 years ago
A ball is being rolled by a normal push of 180N. It is opposed by friction which has a force of 61N and air resistance which has
Nina [5.8K]

Resultant force is basically the force left after everything is added.

if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:

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7 0
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