Which type of reaction is shown below?

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

3 0

3 years ago

Calculate the momentum of a 4 kg Bolling ball being thrown at a speed of 3 m/s

bonufazy [111]

We are given –

Mass of boiling ball is, m = 4 kg
Speed is, v = 3 m/s
Momentum, P =?

As we know –

↠Momentum = Mass × Speed(Velocity)

↠Momentum = 4 × 3 kgm/s

↠Momentum = 12 kgm/s

Henceforth,Momentum will be 12 kgm/s.

7 0

2 years ago

A satellite orbiting the moon very near the surface has a period of110 min. What is free-fall acceleration on the surface of the

Norma-Jean [14]

Answer:

1.54 m/s²

Explanation:

The free-fall acceleration is calculated as

g = w²r

Where w is the angular velocity of the satellite and r is the radius of the moon.

The angular velocity can be calculated as

$w=\frac{2\pi}{T}$

Where T is the period, so

T = 110 min = 110 x 60 s = 6600 s

Then,

$w=\frac{2\pi}{6600\text{ s}}=9.52\times10^{-4}\text{ rad/s}$

Finally, the radius of the moon is r = 1.7 x 10⁶ m, so the free-fall acceleration is

$\begin{gathered} g=w^2r \\ g=(9.52\times10^{-4})^2(1.7\times10^6) \\ g=1.54\text{ m/s}^2 \end{gathered}$

Therefore, the answer is 1.54 m/s²

5 0

1 year ago

A woman drops a vibrating tuning fork, which is vibrating at 513 Hertz, from a tall building. Through what distance (in m) has t

umka21 [38]

Answer:

h = 15.34 m

Explanation:

given,

tuning fork vibration = 513 Hz

speed of sound = 343 m/s

frequency after deflection = 489 Hz

the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases

$f_{apparent} = \dfrac{v}{v+u}f_0$

$489 = \dfrac{343}{343+u}\times 513$

$0.953 = \dfrac{343}{343+u}$

$343+u = \dfrac{343}{0.953}$

$343+u = 359.92$

u = 16.92 m/s

height of the building

v² = u² + 2 g s

16.92² = 2 x 9.8 x h

h = 14.61 m

time taken by sound to reach observer

$t = \dfrac{14.61}{343}$

$t =0.0426\ s$

in this time tuning fork has fallen one more now,

$h' = u t + \dfrac{1}{2}gt^2$

$h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2$

h' = 0.7296 m = 0.73 m

total distance

h = 14.61 + 0.73

h = 15.34 m

6 0

3 years ago

A ball is launched with initial speed v from the ground level up a frictionless hill. the hill becomes steeper as the ball slide

frutty [35]

<span>if we take ball then the Kinetic Energy of the ball is Â˝mVÂ˛ as it rolls up the slope, the Kinetic Energy is converted to potential energy, PE = mgh. mgh = Â˝mVÂ˛ h = VÂ˛/2g</span>

4 0

3 years ago

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