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3 years ago

At what speed would a 3.00 x 10^4 kg airplane have to fly and with a momentum of 1.60 x 10^9 kg.m/s

Physics

1 answer:

Ymorist [56]3 years ago

3 0

Answer:

5.3×10⁴ m/s

Explanation:

From the question,

Momentum = mass× velocity

M = mV................ Equation 1

Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane

make V the subject of the equation

V = M/m.................. Equation 2

Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg

Substitute these values into equation 2

V = 1.6×10⁹/3.0×10⁴

V = 5.3×10⁴ m/s

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Answer:

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3 years ago

What is the difference between Drift current & Diffusion current ?

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3 years ago

A 1 kg object can be accelerated at 10 m/s^2. If you apply this same force to a 4kg object, what will its acceleration be?

AleksAgata [21]

Answer:

$a2 = 2.5~m/s^2$

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force F exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

Assume we apply some given force F to an object of m1=1 Kg that produces an acceleration $a1=10 m/s^2$ , then:

F = m1.a1

The same force F is now applied to a second object m2=4 Kg that produces an acceleration a2, then:

F = m2.a2

Dividing both equations:

$\displaystyle 1=\frac{m1.a1}{m2.a2}$

Solving for a2:

$\displaystyle a2=\frac{m1.a1}{m2}$

Substituting values:

$\displaystyle a2=\frac{1*10}{4}$

$a2 = 2.5~m/s^2$

7 0

3 years ago

A ball, with a mass of 5.9kg, is thrown directly upwards. It reaches a maximum height of 10m from the point at which it was rele

katrin2010 [14]

Answer:

14 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, when the ball is thrown from the ground, it has only kinetic energy, which is given by

$K=\frac{1}{2}mv^2$

where m = 5.9 kg is the mass of the ball and v is its initial speed.

As the ball goes up, its speed decreases, so its kinetic energy decreases and converts into gravitational potential energy. When the ball reaches its maximum height, the speed has become zero, and all the kinetic energy has been converted into gravitational potential energy, given by:

$U=mgh$

where g = 9.8 m/s^2 is the gravitational acceleration and h = 10 m is the maximum height reached by the ball.

Since we can ignore air resistance, energy must be conserved, so the initial kinetic energy must be equal to the final potential energy of the ball, so we can write:

$K=U\\\frac{1}{2}mv^2=mgh$