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Feliz [49]
3 years ago
10

An airplane flies east at a speed of 120 m/s the airplane has a mass of 3200

Physics
1 answer:
EleoNora [17]3 years ago
6 0

1) The momentum of the airplane is 3.84\cdot 10^5 kg m/s east

2) The impulse applied to the airplane is 32,000 kg m/s west

Explanation:

1)

The momentum of an object is defined as

p=mv

where

m is the mass of the object

v is its velocity

The airplane in this problem has

m = 3200 kg is its mass

v = 120 m/s east is its velocity

So, its momentum is

p=(3200)(120)=3.84\cdot 10^5 kg m/s

And since momentum is a vector, it also has a direction, which is the same as the velocity (east).

2)

The impulse applied to an object is equal to the change in momentum of the object:

I=\Delta p = m(v-u)

where

m is the mass

v is the  final velocity

u is the initial velocity

For the airplane in this problem,

m = 3200 kg

u = 120 m/s

v = 110 m/s

So, the impulse is

I=(3200)(110-120)=-32,000 kg m/s

And the negative sign means the direction of the impulse is opposite to the airplane motion, so west.

Learn more about momentum and impulse:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

brainly.com/question/9484203

#LearnwithBrainly

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Nellie pulls on a 10kg wagon with a constant horizontal force of 30N. If there are no other horizontal forces, what is the wagon
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Answer:

The acceleration of the wagon is 3 m/s².

To calculate the acceleration of the wagon, we use the formula below.

Formula:

F = ma............. Equation 1

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F = horizontal Force

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a = acceleration of the wagon.

make a the subject of the equation

a = F/m.............. Equation 2

From the question,

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F = 30 N

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PLEASE HELP 15 POINTS The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. P
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The field lines spread apart as we move away from the charge, and they point away from the charge

Explanation:

The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge at which the field is calculated

There are two pieces of information given by the field lines shown in the graph:

  • The spacing between the lines gives an indication of the strength of the field: the closer to each other they are, the stronger the field. In this case, as we move away from the charge, the spacing between the lines increases, and this means that the field becomes weaker (in fact, it follows an inverse square law, E\propto \frac{1}{r^2}
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3 years ago
Two objects each moving with speed v travel in opposite directions along a straight line passing through both their centers. The
Tpy6a [65]

Answer:

\dfrac{1}{16}

\dfrac{5}{3}

Explanation:

m_1 = Mass of first object

m_2 = Mass of second object

v = Speed of both objects

\dfrac{v}{4} = Combined velocity

The ratio of final kinetic energy to initial kinetic energy will be

\dfrac{K_f}{K_i}=\dfrac{\dfrac{1}{2}(m_1+m_2)(\dfrac{v}{4})^2}{\dfrac{1}{2}(m_1v^2+m_2v^2)} \\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{v^2}{16}}{m_1v^2+m_2v^2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{(m_1+m_2)\dfrac{1}{16}}{m_1+m_2}\\\Rightarrow \dfrac{K_f}{K_i}=\dfrac{1}{16}

The ratio is \dfrac{1}{16}

As the linear momentum is conserved

m_1v-m_2v=(m_1+m_2)\dfrac{v}{4}\\\Rightarrow m_1-m_2=(m_1+m_2)\dfrac{1}{4}

Divide by m_2 on both sides

\dfrac{m_1}{m_2}-1=\dfrac{m_1}{4m_2}+\dfrac{1}{4}\\\Rightarrow \dfrac{m_1}{m_2}-\dfrac{m_1}{4m_2}=\dfrac{1}{4}+1\\\Rightarrow \dfrac{3m_1}{4m_2}=\dfrac{5}{4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5\times 4}{3\times 4}\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{5}{3}

The ratio of mass is \dfrac{5}{3}

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4 years ago
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