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Pani-rosa [81]
3 years ago
6

The Earth orbits the Sun at a speed of 30 km/s. At that speed it completes one path around the Sun every year. Of course, as tha

t happens we are heading toward a different place in space constantly, always accelerating toward the Sun to stay on the circular path. Suppose at a paritcular instant we are headed in the direction of a distant star and we measure the speed of the light from that star as it arrives on Earth. In the vacuum of space we would measurea. 299,792,458 m/sb. 299,792,428 m/sc. 299,792,488 m/sd. 0 m/s
Physics
1 answer:
romanna [79]3 years ago
3 0

Answer:

a. 299,792,458 m/s

Explanation:

Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.

You might be interested in
A uniform beam resting on two pivots has a length L = 6.00 m and weight M = 220 lbs. The pivot under the left end exerts a norma
gulaghasi [49]

Answer:

x = 4,138 m

Explanation:

For this exercise, let's use the rotational equilibrium equation.

Let's fix our frame of reference on the left side of the pivot, the positive direction for anti-clockwise rotation

         ∑ τ = 0

         n₁ 0 - W L / 2 + n₂ 4 - W_woman  x = 0

         x = (- W L / 2 + 4n2) / W_woman

Let's reduce the magnitudes to the SI System

         M = 6 lbs (1 kg / 2.2 lb) = 2.72 kg

         M_woman = 130 lbs = 59.09 kg

Let's write the transnational equilibrium equation

         n₁ + n₂ - W - W_woman = 0

         n₁ + n₂ = W + W_woman

        n₁ + n₂ = (2.72 + 59.09) 9.8

At the point where the system begins to rotate, pivot 1 has no force on it, so its relation must be zero (n₁ = 0)

          n₂ = 605,738 N

 

Let's calculate

         x = (-2.72 9.8 6/2 + 4 605.738) / 59.09 9.8

         x = 4,138 m

4 0
3 years ago
03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
Flauer [41]

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

         w = \sqrt{\frac{k}{m} }

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =\frac{dx}{dt}

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = \frac{dv}{dt}

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

5 0
3 years ago
Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he
storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0
3 years ago
What is the wavelength in air of red light from a helium neon laser?
Elza [17]

Answer:

632.8 nm is the wavelength (in air) of red light from a helium neon laser.

4 0
2 years ago
How Is radar used to determine the speed of a car
Serga [27]
Radars are frequently used to identify distance and speed, such as how far away an object is or how fast it is moving. <span>The </span>radar<span> device can then use the change in frequency to </span>determine the speed<span> at which the </span>car<span> is moving. In laser-</span>speed<span> guns, waves of light are </span>used<span> in place of radio waves.</span>
4 0
3 years ago
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