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irinina [24]
3 years ago
15

How will the gravitational force on a piece of the surface of the star (m1) by the mass of the rest of the star (m2) (effectivel

y located at a point at the center of the star) compare between the AGB and main-sequence phases of a particular star, assuming its mass stays the same?
Physics
2 answers:
NikAS [45]3 years ago
6 0

Answer: The surface will feel a weaker gravitational force during The AGB phase because it farther from the center of the star.

Explanation: The surface will feel a weaker gravitational force during the AGB phase because it farther from the center of the star. Surface mass loss via a wind during the AGB phase and/or via catastrophic ... of heat, pressure, and gravitational diffusion are ignored.

Anna [14]3 years ago
6 0

Answer:

Answer: Gravitational weak force will be felt on the surface during the asymptotic giant branch phase because it is far away from the middle of the star.

Explanation:

(AGB) which is know as asymptotic giant branch is a space on the Hertzsprung–Russell drawing which been populated by cool luminous stars which are evolved.

When the surface will be felt to be weak, is during the period of asymptotic giant branch(AGB) phase, which will be very far away from the middle of the star.

The will be loss of the surface mass through a wind during the AGB phase and/or through catastrophic of gravitational diffusion, pressure, heat are been left out.

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En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
A 15-lb block B starts from rest and slides on the 25-lb wedge A, which is supported by a horizontal surface. Neglecting frictio
xxTIMURxx [149]

The angle of the wedge is 30°.

Answer:

5.88 ft/s

Explanation:

a) The block will slide down due to it's weight.

initial velocity u= 0

final velocity, v

acceleration, a = g sin 30° = 32 ft/s²× sin 30° = 16 ft/s²

Sliding displacement, s = 3ft

Use third equation of motion:

v^2-u^2 = 2as

substitute the values and solve for v

v^2-0 = 2\times 16 \times 3 =96 ft^2/s^2\\v = 9.8 ft/s

b) Use conservation of momentum:

Initial momentum of the system  = 0

final momentum = (15) ( 9.8)+ (25)(v')

v' = 5.88 ft/s

3 0
3 years ago
Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where
fiasKO [112]

The emf induced in the second coil is given by:

V = -M(di/dt)

V = emf, M = mutual indutance, di/dt = change of current in the first coil over time

The current in the first coil is given by:

i = i₀e^{-at}

i₀ = 5.0A, a = 2.0×10³s⁻¹

i = 5.0e^(-2.0×10³t)

Calculate di/dt by differentiating i with respect to t.

di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

V = 43 volts

3 0
3 years ago
How did mount st. Helen get its name
steposvetlana [31]

The modern name, Mount St. Helen's, was given to the volcanic peak in 1792 by seafarer and explorer Captain George Vancouver of the British Royal Navy. He named it in honor of fellow countryman Alleyne Fitzherbert, who held the title 'Baron St. Helen's.

5 0
3 years ago
Convertir 1200 ms a cs<br> Convertir 0,3 mm a cm.
Vaselesa [24]

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

3 0
3 years ago
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