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3 years ago

Degeneracy pressure stops the crush of gravity in all the following except

Physics

1 answer:

quester [9]3 years ago

5 0

Answer:

A very massive main- sequence star

Explanation:

Degeneracy pressure refers to pressure expend by dense material , which is composed of fermions, which is an example of electrons in a white dwarf star. According to Pauli exclusion principle, which states that no two fermions can exist in the same quantum state, actually give details about the pressure.

When there is stellar masses that is less than about 1.44 of that of solar masses, there will be gravitational fall in the energy , and the energy will not be sufficient to produce the neutrons of a neutron star, therefore, the fall is abstruptly stopped through the electron degeneracy to form white dwarfs.this result to the creation of an effective pressure that further prevents gravitational fall.

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an object is moving in a straight line with a constant acceleration initially it is traveling at 16 meters per second. 3 seconds

VLD [36.1K]

A)It moved 6 meters.

B)It would 3 meters.

I think this is the answer.

Have a good day.

4 0

3 years ago

A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from

NikAS [45]

The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

distance traveled by the car, s = 1.2 m
time of motion of the car, t = 0.67 s
initial velocity of the car, u = 0

The acceleration of the car is calculated as;

$s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2$

The final velocity of the toy car is calculated as;

$v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s$

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

Learn more here: brainly.com/question/20352766

6 0

2 years ago

A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

So, the energy dissipated during the skidding of car:

Frictional force:

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

Now from the work-energy equivalence:

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

3 years ago

Please help me i have this due tommorrow!!!

Ronch [10]

I found this!! maybe this will help :)

8 0

2 years ago

A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,

egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0

3 years ago