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Sergio [31]
3 years ago
14

What is the acceleration of an object that has a mass of 10kg and is pushed with a force of 50n

Physics
1 answer:
worty [1.4K]3 years ago
7 0

Answer:

5m/s/s

Explanation:

force = mass x acceleration

50 = 10a

a=5m/s/s

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One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤
OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

P(t)=60 W, 0\leq t

P(t)=25 W, 5.0\leq t

Aplying the values to the equation we have:

Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}

Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}

Power_{avg} = 42.5W

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4 years ago
A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
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Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

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