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vlada-n [284]
2 years ago
6

Owen and Dina are at rest in frame S' , which is moving at 0.600 c with respect to frame S . They play a game of catch while Ed

, at rest in frame S , watches the action (Fig. P39.75). Owen throws the ball to Dina at 0.800 c (according to Owen), and their separation (measured in S' ) is equal to 1.80 × 10¹²m .(a) According to Dina, how fast is the ball moving?
Physics
1 answer:
Taya2010 [7]2 years ago
6 0

The speed of the ball with respect to Dina is 0.800c.

A frame of reference is a set of reference points—geometric points whose positions are known mathematically and physically—that define the origin, orientation, and scale of an abstract coordinate system.

If a body does not continuously adjust its position in relation to its environment throughout the course of time, it is said to be at rest.

In frame S', Dina and Owen are at rest.

The speed of the ball with respect to Owen, u =0.800c

The speed of the fram S' with respect to frame S,  v = 0.600c

Distance between Dina and Owen, L(p) = 1.8 × 10¹² m

Speed of light, c = 3 × 10⁸ m/s

Therefore, the speed of the ball according to Dina is 0.800c. As Dina and Owen are in the same frame.

Learn more about the frame of reference here:

brainly.com/question/10962551

#SPJ4

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Answer:

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Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp
Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

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A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
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Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

\theta=10.47^{\circ}

Hence, this is the required solution.

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