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WINSTONCH [101]

3 years ago

5

forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s

² east. Determine the magnitude of F1 and F2.

1 answer:

zhenek [66]3 years ago

5 0

Answer:

345453-5676

its the right answer

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Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star.

Vsevolod [243]

Answer:

The angular speed of the neutron star is 3130.5 rad/s.

Explanation:

Given that,

Initial radius $r_{1}=7\times10^{5}\ km$

Final radius $r_{2}=18 km$

Density of a neutron $\rho= 10^{14}$

Equal masses of two stars $m_{1}=m_{2}$

Suppose, If the original star rotated once in 35 days, find the angular speed of the neutron star

Time period of original star T = 35 days = 3024000 s

We need to calculate the initial angular speed of original star

Using formula of angular star

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega_{1}=\dfrac{2\pi}{3024000}$

$\omega_{1}=0.00207\times10^{-3}\ rad/s$

Let the initial moment of inertia of the star is

$I_{1}=m_{1}r_{1}^2$

Final moment of inertia of the star is

$I_{2}=m_{2}r_{2}^2$

From the conservation of angular momentum

$I_{1}\omega_{1}=I_{2}\omega_{2}$

$\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}$

$\omega_{2}=\dfrac{m_{1}r_{1}^2\omega_{1}}{m_{2}r_{2}^2}$

Put the value into the formula

$\omega_{2}=\dfrac{(7.0\times10^{5})^2\times0.00207\times10^{-3}}{18^2}$

$\omega_{2}=3130.5\ rad/s$

Hence, The angular speed of the neutron star is 3130.5 rad/s.

8 0

3 years ago

A train with mass 3.3 x 107 kg starts from rest and accelerates to a speed of 42

Natasha2012 [34]

Answer:

kinetic energy of the train = 2,910.6 x 10⁷ joule

Explanation:

Given:

Mass of train = 3.3 x 10⁷ kg

Speed of train = 42 m/s

Find:

kinetic energy of the train

Computation:

kinetic energy = (1/2)(m)(v²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)

kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)

kinetic energy of the train = (3.3 x 10⁷)(882)

kinetic energy of the train = 2,910.6 x 10⁷ joule

4 0

3 years ago

Read 2 more answers

A 104.328-kg baserunner slides into home base. He starts his slide moving at 8.451 m/s and 6.596 m from home base; when he cross

JulijaS [17]

The coefficient of kinetic friction between the baserunner and the ground is 0.51.

<h3>Coefficient of kinetic friction </h3>

The coefficient of kinetic friction between the base runner and the ground is calculated as follows;

μ = a/g

where;

a is acceleration
g is gravity

v² = u² + 2as

a = (v² - u²)/(2s)

a = (8.451² - 2.322²)/(2 x 6.596)

a = 5 m/s²

μ = 5/9.8

μ = 0.51

Thus, the coefficient of kinetic friction between the baserunner and the ground is 0.51.

Learn more about coefficient of friction here: brainly.com/question/14121363

#SPJ1

7 0

2 years ago

Larry sees a group of people weeping, with frowns on their faces and their eyes turned down. Larry their expressions to understa

Salsk061 [2.6K]

Answer:

c

Explanation:

no need explanation u can trust me

8 0

3 years ago

At the start of a hockey game, the referee drops the puck between two players from opposing teams. Each player wants to push the

spin [16.1K]

Answer:

a) puck is subjected to both the forces of the hockey sticks in a horizontal direction,

b)the puck does not move since the sum of the forces is zero

c) changing the magnitude or direction of its applied force

Explanation:

a) The puck is subjected to both the forces of the hockey sticks in a horizontal direction, these forces are of equal magnitude and opposite direction since the puck is at rest.

In the direction of the y-axis (perpendicular to the ice) you have the weight of the disk and the normal to this weight that are also in equilibrium.

b) the puck does not move since the sum of the forces is zero, which implies that the forces of the hockey sticks are of equal magnitude and opposite direction.

c) the player has several ways to make the puck move

* slightly changing the angle of the club and therefore the direction of the force, in this case the disc comes out in the direction of this component

* inclined the stick slightly so that the force has a vertical component and the puck jumps in this direction

* Increasing the magnitude of the force so that the puck comes out in the opposite direction to the player

* The worst case, decreasing its force to zero and the disk comes out in its direction by the other force that had the same magnitude.

3 0

3 years ago