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iren [92.7K]
3 years ago
6

A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass

of 2.78×1030 kg and the black hole has a mass of 9.94×1030 kg. What is the magnitude of the gravitational attraction between the two?
Physics
1 answer:
m_a_m_a [10]3 years ago
5 0

Answer:

  F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

F=G\dfrac{m_1m_2}{R^2}

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
what is the name of the tool that allows you to copy two or more shapes into a single part? when working with 3D
Anna [14]

Answer:

Modeling tool or Align tool. it depends what type of sandbox platform you use

Explanation:

1

8 0
3 years ago
A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is
Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d=\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m

Wavelength of light,\lambda=510 nm=510\times 10^{-9} m

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

sin\theta=\frac{m\lambda}{d}

For first bright fringe, =1

sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305

\theta=sin^{-1}(0.305)=17.76^{\circ}

The distance between two m=1 fringes

x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

sin\theta=1

sin\theta=\frac{m\lambda}{d}

Substitute the values

1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}

m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3

Maximum number of bright fringes on the scree=2m+1=2(3)+1=7

8 0
3 years ago
I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi
Masteriza [31]

Answer:

<h3>a.</h3>
  • After it has traveled through 1 cm : I(1 \ cm) = 0.5488 I_0
  • After it has traveled through 2 cm : I(2 \ cm) = 0.3012 I_0
<h3>b.</h3>
  • After it has traveled through 1 cm : od( 1\ cm) =  0.2606
  • After it has traveled through 2 cm :  od( 2\ cm) =  0.5211

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient \mu the formula is:

I(x) = I_0 e^{-\mu x}

where I is the intensity of the beam, I_0 is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = I_0 e^{- 0.6}

I(1 \ cm) = 0.5488 \ I_0

After travelling 2 cm:

I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = I_0 e^{- 1.2}

I(2 \ cm) = 0.3012 \ I_0

<h2>b</h2>

The optical density od is given by:

od(x) = - log_{10} ( \frac{I(x)}{I_0} ).

So, after travelling 1 cm:

od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )

od( 1\ cm) = - log_{10} ( 0.5488 )

od( 1\ cm) = - (  - 0.2606)

od( 1\ cm) =  0.2606

After travelling 2 cm:

od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )

od( 2\ cm) = - log_{10} ( 0.3012 )

od( 2\ cm) = - (  - 0.5211)

od( 2\ cm) =  0.5211

3 0
3 years ago
In order for an object to accelerate, a forcegravityfrictiona swing must be applied.
vodomira [7]
<span>In order for an object to accelerate, a <u>force</u> must be applied. It follows Newton’s second law of motion where it states that a body at rest remains at rest unless a force is acted upon it. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. Keep in mind that when you exert work, you are exerting energy too. </span>
4 0
3 years ago
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