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Volgvan
3 years ago
12

Astronomers discover a bright X-ray source in the Milky Way. They see no counterpart in the optical bands, but it has another st

ar orbiting it, and they determine that the X-ray source has a mass of 10 MSun. What is the most likely origin of the X-rays?
a. a black hole
b. a neutron star
c. a white dwarf
d. a gamma-ray burst
Physics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

a. a black hole

Explanation:

X-ray emission from the central degrees of the Milky Way Bright X-ray emission traces the coherent edge brightened shell-like feature, dubbed the northern chimney, located north of Sgr A* and characterized by a diameter of about 160 pc. On the opposite side, the southern chimney appears as a bright linear feature. Bright X-ray emission is observed at high latitude

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A wire, of length L = 4.1 mm, on a circuit board carries a current of I = 1.96 μA in the j direction. A nearby circuit element g
tamaranim1 [39]

Answer:

The magnitude of the magnetic field B is 5.921 T.

Explanation:

Given that,

Length = 4.1 mm

B_{x}=4.9\ G

B_{y}=2.3\ G

B_{z}=2.4\ G

Current I = 1.96\ mu A

We need to calculate the magnetic field

Using formula of magnetic field

B=\sqrt{B_{x}^2+B_{y}^2+B_{z}^2}

Put the value into the formula

B=\sqrt{(4.9)^2+(2.3)^2+(2.4)^2}

B=5.921\ T

Hence, The magnitude of the magnetic field B is 5.921 T.

6 0
3 years ago
Caleb is filling up water balloons for the Physics Olympics balloon tosscompetition. Caleb sets a 0.50-kg spherical water balloo
Mashcka [7]

a)

• P = F/A

P = pressure = 630 N/m^2

F = force

A = area

F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N

m= mass

g= gravity

P = F/A

A = F/P

A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2

b)

• Area of a circle = pi* radius ^2

7.778 x 10^-3 m^2 = pi* radius ^2

√(7.778 x 10^-3 m^2 / pi ) = radius

radius = 0.04976 m

Answers:

a ) 7.778 x 10^-3 m^2

b) 0.04976 m

8 0
1 year ago
.A coin rolls off the edge of a table. The coin
geniusboy [140]

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

s  = ut +  \frac{1}{2}a {t}^{2}   \\ 0.8 = 1.7 \times t \\  \frac{0.8}{1.7}  = t \\ t = 0.47s

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

s = ut +  \frac{1}{2} a {t}^{2}  \\ s =  \frac{1}{2}  \times 10 \times (0.47) ^{2}  \\ s = 1.1045m

5 0
2 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
Add a small ball to a graduated cylinder containing 10 milliliters of water.
malfutka [58]
The volume of water will increase . If yu subtract the original volume from the new volume of water you will get the volume of the small ball.
8 0
3 years ago
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