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malfutka [58]
2 years ago
11

Which is a valid inertial frame of reference?

Physics
1 answer:
kow [346]2 years ago
6 0

The correct answer to the question will be C. a hot air balloon moving at 30° east of north with no net force

EXPLANATION:

Before coming into any conclusion, first we have to understand an inertial frame of reference.

An inertial frame of reference is a frame of reference which is an unaccelerated frame. In this frame of reference, Newton's laws are valid.

A falling rock can not be considered as an inertial frame as it is accelerating. When a body falls, the forces that act on the body are gravity and air resistance. The gravity acts in vertical downward direction while air resistance acts in vertical upward direction. There is a net force in the downward direction which means it is experiencing an acceleration.

The frictionless spinning merry-go-round can not be considered as an inertial frame of reference. It is so because a rotating frame can not be considered as an inertial frame. It is non inertial in nature.

A hot air balloon moving 30 degree east of north with no net force can be considered as an inertial frame as it is not accelerating.

A space shuttle whose boosters just ignited for take off can not be also inertial frame as it is an accelerated frame.

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I think the correct answer is true. Most paper does not reflect light very well because its surface is somewhat rough. Light only reflects to surfaces which has a smooth texture or have a uniform texture on the surface. Hope this answers the question. Have a nice day.
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The electric current in a copper wire is composed of what
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Answer:

A copper wire current consists of electrons appropriately called conduction electrons.

Explanation:

This answer came from quizlet.com. I hope that this helps you and good luck!

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2 years ago
Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

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\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

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2 years ago
At the intersection of Texas Avenue and University Drive,
Zielflug [23.3K]

Answer:

  • The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s  

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector \hat{i} pointing north and \hat{j} pointing east, the final velocity will be

\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )

\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

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\vec{P}_f = (m_{car}+ m_{truck}) * V_f

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\vec{P}_f = (2.850 \ kg \ ) *  ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )

\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )

As there are not external forces, the total linear momentum must be constant.

So:

\vec{P}_0= \vec{P}_f

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\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} ) 

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\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}  } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.

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m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}

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v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}

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950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}

v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}

v_{car}=19.524 \frac{m}{s}

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