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4 years ago

Discuss the limitations of using the Doppler shift to determine an object's speed.

Physics

1 answer:

pantera1 [17]4 years ago

3 0

Answer and Explanation:

Limitation of Doppler shift :

The Doppler impact is relevant when the speeds of the wellspring of sound and spectator are considerably less than the speed of sound. The movement of both the spectator and the source is along a similar straight line.When movement is not in straight line or velocity is not much less than speed of light then we can not use Doppler shift

This is the limitation of Doppler shift to determine the object distance

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Can someone please help meee .

fiasKO [112]

Answer:

32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice

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3 years ago

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You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

 $t = 1.05s$ 

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3 years ago

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

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3 years ago

The kinetic energy of a body of mass 15 kg is 30 joule. What is its momentum?

lys-0071 [83]

This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:

Kinetic energy = (1/2) · (mass) · (speed²)

Momentum = (mass) · (speed)

So, now ... We know that

==> mass = 15 kg, and

==> kinetic energy = 30 Joules

Take those pieces of info and pluggum into the formula for kinetic energy:

Kinetic energy = (1/2) · (mass) · (speed²)

30 Joules = (1/2) · (15 kg) · (speed²)

60 Joules = (15 kg) · (speed²)

4 m²/s² = speed²

Speed = 2 m/s

THAT's all you need ! Now you can find momentum:

Momentum = (mass) · (speed)

Momentum = (15 kg) · (2 m/s)

Momentum = 30 kg·m/s

(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)

8 0

3 years ago

The angular speed of a rotating platform changes from

Leviafan [203]

Answer:

Explanation:

The angular speed of a rotating platform changes from.at a constant rate as the platform moves through an angle

3 0

3 years ago