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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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An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos
erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v}    \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm

3 0
3 years ago
At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp
motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

K = \frac{1}{2} mv^2

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

5 0
3 years ago
You must record your daily workouts on a
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I do a yoga mat that could be different to your answer

5 0
3 years ago
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A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds
Nesterboy [21]

Answer:

c. 4 meters/second

Explanation:

The formula to calculate average speed is:

s = \frac{x_{2} - x_{1}  }{ t_{2} - t_{1}  }

7 0
3 years ago
g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a
zimovet [89]

Answer:

 λ = 3 10⁻⁷ m,   UV laser

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

let's use trigonometry

         tan θ = y / L

as in this phenomenon the angles are small

        tan θ = \frac{sin \ \theta}{cos \ \theta} = sin θ

        sin θ = y / L

we substitute

      a y / L = m  λ

let's apply this equation to the initial data

       a  0.04 / L = 1 600 10⁻⁹

       a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

      a 0.04 / L = 2  λ

       a / L = 50  λ

we solve the two expression is

         1.5 10⁻⁵ = 50  λ

          λ = 1.5 10⁻⁵ / 50

          λ = 3 10⁻⁷ m

    UV laser

3 0
3 years ago
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