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lina2011 [118]
2 years ago
14

Viewed from a distance, how would a flashing red light appear as it fell into a black hole?.

Physics
1 answer:
Pachacha [2.7K]2 years ago
5 0

Explanation:

Light rays that pass close to the black hole get caught and cannot escape. so what ends up happening to the light is that it starts bending due to the strong gravitational force of the black hole.

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Three identical lights are connected in series to a 12V battery. How does the brightness of each light compare?
kodGreya [7K]

Answer: They are identical brightness

Explanation:

If the lights are assumed to be resistance bulbs

Each light has the same current and will each drop one third of the supply voltage.  

8 0
3 years ago
A cyclist maintains a constant velocity of
tensa zangetsu [6.8K]

Hi there!

We can use the equation:

d = x₀ + vt, where:

x₀ = initial distance from the reference point

v = velocity (m/s)

t = time (sec)

Plug in the given values:

d = 248 + 5(49)

d = 493m

6 0
3 years ago
The weight of a girl with a mass of 40.0 kg is_N
faltersainse [42]

Answer: The weight of a girl with a mass of 40kg is 392.266 Newtons.

Explanation:

5 0
3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
If i fail choir do i fail 7th grade
Strike441 [17]

Answer:

no ....................

7 0
3 years ago
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