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3 years ago

A neutron star is moving in outer space at 4,000 km/hr. What happened to the star that set it in motion on it's current course?

Physics

1 answer:

astra-53 [7]3 years ago

4 0

Answer:

a balanced force acted on it and propelled it to 4,000 km/hr

Explanation:

For the neutrons star which is moving in outer space at 4,000 km/hr, it could only be possible as a result of the balanced force which had already acted on it. <em>This is based on newton's law of motion which states that 'To every action, there is equal and opposite reaction'. </em>

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A centrifuge in a medical laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates through 52.0 re

eduard

The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

<h3> Constant angular acceleration</h3>

Apply the following kinematic equation;

ωf² = ωi² - 2αθ

where;

ωf is the final angular velocity when the centrifuge stops = 0
ωi is the initial angular velocity
θ is angular displacement
α is angular acceleration

ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s

θ = 52 rev x 2π rad/rev = 326.7 rad

0 = ωi² - 2αθ

α = ωi²/2θ

α = ( 356.05²) / (2 x 326.7)

α = 194.02 rad/s²

Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².

Learn more about angular acceleration here: brainly.com/question/25129606

#SPJ1

7 0

2 years ago

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 49.8 N,

tatiyna

Answer:

W = 7.06 J

Explanation:

From the given information the spring constant 'k' can be calculated using the Hooke's Law.

$F = kx\\49.8 = k(0.181)\\k = 275.13~N/m$

Now, using this spring constant the additional work required by F to stretch the spring can be found.

The work energy theorem tells us that the work done on the spring is equal to the change in the energy. Therefore,

$W = U_2 - U_1\\W = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 = \frac{1}{2}(275.13)[0.29^2 - 0.18^2] = 7.06~J$

6 0

3 years ago

An object has a mass of 10.2 kg, what is its weight in Newton's?

Juliette [100K]

Answer:

The answer is 98 N

Explanation:

I am pretty sure

3 0

2 years ago

Read 2 more answers

Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i

Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

$L = L_i_n_i + L_3_s$

$L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt$

$L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s$

Moment if inertia is

$I = 2ml^2$

$I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2$

Angular speed

ω = L/I

$= 38 / 5.75\\\\=6.61$

The speed of each ball is

V = ωL

$= 6.61\times0.5\\\\= 3.31m/s$

7 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago