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s344n2d4d5 [400]
4 years ago
6

Are there really dead stars in the sky?​

Physics
2 answers:
sp2606 [1]4 years ago
5 0

Answer:

Yes

Explanation:

The star would have died over thousands of years ago but the light takes a long time for it to reach us on Earth.

balu736 [363]4 years ago
4 0

Answer:

Eventually the cycle of star birth and death will come to an end. Gravity will have won, a victory delayed by the ability of stars to call on the resources of nuclear fusion. But ultimately, gravity will reduce all stars to a super-dense state as black holes, neutron stars or cold white dwarfs.

Explanation:

You might be interested in
Are you a negro can i call you one
Roman55 [17]

Answer:

no it is mean call people their names

Explanation:

because if my name was Shar i would want people to call me Shar not negro

3 0
3 years ago
How can you change the phases of matter using thermal energy
Leokris [45]

Answer:

These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance.

8 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi
steposvetlana [31]

Answer:

W=K_f-K_i

Explanation:

The work done on a particle by external forces is defined as:

W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,

According to Newton's second law F=ma. Thus:

W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\

Acceleration is defined as the derivative of the speed with respect to time:

W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,

Speed is defined as the derivative of the position with respect to time:

W=m\int\limits^{v_f}_{v_i} v \cdot dv \,

Kinetic energy is defined as K=\frac{mv^2}{2}:

W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i

3 0
3 years ago
An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh
melomori [17]

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.


7 0
3 years ago
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