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3 years ago

What are the two forces that oppose each other while a star?

1 answer:

5 0

While self-gravity pulls the star inward and tries to make it collapse, thermal pressure (heat created by fusion) pushes outward. These two forces cancel each other out in a main sequence star, thus making it stable.

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A car going 50 miles/hr accelerates to pass a truck. 5 seconds later the car is going 80 miles/hr. What is the acceleration of t

mariarad [96]

It would be 6 mph<span>/s 11</span>

5 0

3 years ago

I don't understand help

Bumek [7]

Answer:

Pe is stored energy to do work Pe is stored in objects that are at rest

KE defined as the energy of motion

began to move

Not sure about the rest

hope this helps

Explanation:

6 0

3 years ago

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

$m = 25 * 12.5$

$m \approx 310$

6 0

3 years ago

Please refer to the picture

Vinvika [58]

I think is c not sure though

3 0

3 years ago

A 0.05-m3 rigid tank initially contains refrigerant-134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to

lesya692 [45]

Answer:

A= 203 KJ

B= 54 Kg

Explanation:

The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:

Δm = m₂ – m₁

= V(1/α₂ – 1/α₁)

= 0.05 (1/0.0008935 – 1/ 0.025645)Kg

= 54Kg

The heat transfer is obtained from the energy balance:

ΔU= $m_i_n$ $h_i_n$ + $Q_n_e_t$

m₂u₂ – m₁u₂ = $m_i_n$ $h_i_n$ + $Q_n_e_t$

$Q_n_e_t$ = m₂u₂ – m₁u₁ – $m_i_n$

= V/α₂u₂ - V/α₁u₁ – $m_i_n$

=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj

= 203 KJ

8 0

3 years ago