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noname [10]
3 years ago
14

Antiballistic missiles (ABMs) are designed to have very large accelerations so that they may intercept fast-moving incoming miss

iles in the short time available. What is the takeoff acceleration of a 10,000-kg ABM that expels 196 kg of gas per second at an exhaust velocity of 2.50×103 m/s?
Physics
1 answer:
4vir4ik [10]3 years ago
3 0

Answer:

a = 39.2 m/s²

Explanation:

The <em>takeoff acceleration </em>of a missile or a rocket can be expressed by the equation

  • a = V₀/m * Δm/Δt - g

Where V₀ is the exhaust velocity, m is the mass of the missile, Δm is the mass of gas expelled during a timelapse Δt, and g is gravitational acceleration.

All required data is given by the problem, so now we <u>solve for a</u>:

a = 2.50x10³m/s ÷ 10,000kg * 196kg ÷ 1s - 9.8m/s²

a = 39.2 m/s²

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On an ice skating rink, a girl of mass 50 kg stands stationary, face to face with a boy of mass 80 kg. The children push off of
sasho [114]

The velocity of the girl is  -4.8 m/s.

Using the principle of conservation of linear momentum, The total momentum of  bodies before and after collision is constant. Since the two objects are stationary, the initial momentum of each body is zero.

Thus;

0 = (80 × 3) + (50 × v)

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v = -4.8 m/s

Note that the negative sign shows that the velocity of the girl is in opposite direction that that of the girl.

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3 years ago
Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (
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Answer:

a) When moving towards a high pressure center the pressure values ​​increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values ​​85.5 kPa.

When moving towards a high pressure center the pressure values ​​increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

6 0
3 years ago
1. Aunt Minnie gives you $10 per second for 4 seconds. How much money do you have after 4 se
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Answer:

$40

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10 × 4 = 40

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4 0
3 years ago
Read 2 more answers
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
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