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xenn [34]
3 years ago
6

If a black hole itself emits no radiation, what evidence do astronomers and physicists today have that the theory of black holes

is correct?
Physics
1 answer:
AleksandrR [38]3 years ago
6 0

Answer:

Einstein's general theory of relativity is the theory behind black holes has been tested with a wide range of experiments of which all confirm the predictions the theory makes. We cannot see black holes phenomena inside the event horizon, we do observe things outside this limit.

Black holes in binary star systems leave signs of their presence on neighboring star thats detected and the signs include X-ray emissions, accretion disks, and large orbit perturbations.

this is the evidence that astronomers and physicists have to show that the theory about black holes is correct.

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An ax is an example of a ____________.
natta225 [31]
<span>An ax is an example of a wedge. The correct option among all the options that are given in the question is the second option or option "b". The other choices given in the question are incorrect and can be easily neglected. I hope that this is the answer that has actually come to your great help.</span>
4 0
3 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
3 years ago
Which of the following statements are true?
Likurg_2 [28]

Answer:

Explanation:

(A) True: It is true.

In junction law, the current entering at a junction is equal to teh current leaving at the junction.

(B) False: It is false.

The kirchhoff's junction law is based on the conservation of charge.

(C) True: It is true.

Energy is used in the circuit.

(D) True: It is true.

It is based on the conservation of charge.

4 0
3 years ago
Answer it pls!!!!!!!!!!!
Archy [21]

Answer:

Fractional error = 0.17

Percent error = 17%

F = 112 ± 19 N

Explanation:

Plug in the values to find the force:

F = (3.5 kg) (20 m/s)² / (12.5 m) = 112 N

Find the fractional error:

ΔF/F = Δm/m + 2Δv/v + Δr/r

ΔF/F = 0.1/3.5 + 2(1/20) + 0.5/12.5

ΔF/F = 0.17

Multiply by 100% to find the percent error:

ΔF/F × 100% = 17%

Solve for the absolute error:

ΔF = 0.17 × 112 N = 19 N

Therefore, the force is:

F = 112 ± 19 N

8 0
3 years ago
A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove
jonny [76]

We must remember that the total net force equation at constant velocity is:

<span>F – Ff = 0</span>

of

F - µN = 0

Using Newton's 2nd Law of Motion:<span>

F = m a 

<span>Where,

F = net force acting on the body 
m = mass of the body 
a = acceleration of the body 

Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to 

F = net Force = 0 

Therefore, 

F - µN = 0 

where 

µ = coefficient of friction = 0.20 
N = normal force acting on the cart = 12 N 

Therefore, 

F - 0.20(12) = 0 

<span>F = 2.4 N </span></span></span>
4 0
3 years ago
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