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dusya [7]
2 years ago
7

For general projectile motion, which of the following best describes the horizontal component of a projectile's acceleration? As

sume air resistance is negligible.
-The horizontal component of a projectile's acceleration continually decreases.
-The horizontal component of a projectile's acceleration is zero.
-The horizontal component of a projectile's acceleration remains a nonzero constant.
-The horizontal component of a projectile's acceleration continually decreases.
-The horizontal component of a projectile's acceleration initially decreases and then increases.
Physics
2 answers:
butalik [34]2 years ago
7 0

Answer:

The horizontal component of a projectile's acceleration is zero

Explanation:

When calculating all the different components of a projectile (acceleration, initial velocity, final velocity, etc.), you need to break it up into its X and Y directions.

During free fall, the only force acting on the projectile is gravity, and since gravity acts down (in the Y direction), there is no force acting in the X direction. Hence, the acceleration in the X direction is zero.

ANEK [815]2 years ago
6 0
The horizontal component of a projectile acceleration is zero
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If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
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Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

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   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

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Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

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3 0
2 years ago
7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has
NNADVOKAT [17]

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The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

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The mass of water per second that strikes the blade is, \dfrac{m}{t}=30\ kg/s

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Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

F=\dfrac{\Delta P}{\Delta t}

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F=30\ kg/s\times (-16-16)\ m/s

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0
2 years ago
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