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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?

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 65 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5 m/s. how fast is he going as he l

KIM [24]

m = mass of trampoline artist = 65 kg

v₀ = initial speed of the artist at the top of platform = 5 $\frac{m }{s}$

h = height through which the artist drop before landing on trampoline = 3 m

v = final speed of the artist just before landing on trampoline = ?

using conservation of energy

Kinetic energy of artist just before landing = initial kinetic energy at the top of platform + potential energy at the top of platform

(0.5) m v² = (0.5) m v₀² + mgh

dividing each term by "m"

(0.5) v² = (0.5) v₀² + gh

inserting the values

(0.5) v² = (0.5) (5)² + (9.8)(3)

v = 9.2 $\frac{m }{s}$

x = compression of the trampoline = ?

k = spring stiffness constant = 62000 $\frac{N }{m}$

Assuming the lowest depression point as reference line for measuring the potential energy

using conservation of energy

kinetic energy of artist + potential energy of artist before landing = spring potential energy of trampoline

(0.5) m v² + mg x = (0.5) k x²

inserting the values

(0.5) (65) (9.2)² + (65 x 9.8) x = (0.5) (62000) x²

x = 0.31 m

6 0

3 years ago

A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves h

Marina CMI [18]

Answer:

The thickness is $t = 0.5615 \ cm$

Explanation:

From the question we are told that

The wavelength of the of the rader waves is $\lambda = 2.92 \ cm$

The index of refraction of the polymer is $n = 1.30$

The thickness is mathematically represented as

$t = \frac{\lambda }{4 n }$

Substituting values

$t = \frac{2.92}{4 * 1.30 }$

$t = 0.5615 \ cm$

6 0

3 years ago

Something is thrown in the air and 5.35 seconds later it lands on something 2.1 meters from the ground. How fast was it thrown u

UNO [17]

Answer:

26.6 m/s

Explanation:

Given:

Δy = 2.1 m

t = 5.35 s

a = -9.8 m/s²

Find: v₀

Δy = v₀ t + ½ at²

(2.1 m) = v₀ (5.35 s) + ½ (-9.8 m/s²) (5.35 s)²

v₀ = 26.6 m/s

5 0

3 years ago

Which of the following is the brightness of a star as we see if from Earth?

My name is Ann [436]

I think that it is apparent magnitude

8 0

3 years ago

A ball thrown vertically upward is caught by the thrower after 4.00 sec. Find the initial velocity of the ball and the maximum h

My name is Ann [436]

Answer:

Initial velocity = 39.2m/s

Maximum height is 78.4m

Explanation:

Given

$Time, t = 4s$

Solving (a): Initial Velocity

Using first law of motion:

$v = u + at$

Where

$v = final\ velocity = 0$

$u = iniital\ velocity = ??$

<em /> $a = acceleration = -g$ <em> [g represents acceleration due to gravity]</em>

$t = 4$

Substitute these value in the above formula:

$v = u + at$

$0 = u - g * 4$

$0 = u - 9.8 * 4$

Take g as 9.8m/s²

$0 = u - 39.2$

$u = 39.2m/s\\$

<em>Hence, initial velocity = 39.2m/s</em>

Solving (b): Maximum Height

This will be solved using second equation of motion

$s = ut + \frac{1}{2}at^2$

This becomes

$s = ut - \frac{1}{2}gt^2$

Substitute values for u, t and g

$s = 39.2 * 4 - \frac{1}{2} * 9.8 * 4^2$

$s = 156.8 - 78.4$

$s = 78.4$

<em>Hence, the maximum height is 78.4m</em>

7 0

4 years ago