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MrRa [10]
3 years ago
12

Intro/forces honors physics please help asap

Physics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

I think it should be D

Explanation:

This is because the arrow pointing right is bigger than the arrow pointing left. Hence we can conclude the the force acting on the right is greater and hence it is accelerating

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I think the term is a variable?
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A turtle moves at an average speed of 0.2m/S towards the finish line that is 80m away from the starting line. A rabbit,racing wi
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2 years ago
Max and Jimmy want to jump on a trampoline. Max begins jumping in a steady pattern, making small waves in the trampoline. Jimmy
mylen [45]

Answer:

x_total = (A + B) cos (wt + Ф)

we have the sum of the two waves in a phase movement

Explanation:

In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.

         Max

               x₁ = A cos (wt + Ф)

         Jimmy

              x₂ = B cos (wt + Ф)

         

total movement

             x_total = (A + B) cos (wt + Ф)

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8 0
3 years ago
A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red
Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

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Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }       ......(1)

Here \lambda_{s} is source wavelength, \lambda_{o} is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, \lambda_{s} = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, \lambda_{0} = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }

\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84

{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056

1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}

\frac{v}{c}=\frac{0.2944}{8.056}

v = 0.036c=0.036\times3\times10^{8}

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