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3 years ago

How do you think the formation of new stars is related

Physics

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

when the matter from old stars reform

Explanation:

You might be interested in

Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?

Ainat [17]

PV = 400 x 0.08 = 32 J

Hope this helps

7 0

3 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago

Convert 0.700 atm of pressure to its equivalent in millimeters of mercury.

inna [77]

Answer:

532 millimeters of mercury

Explanation:

In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:

1 atm = 760 mm Hg

Therefore, we can solve the problem by setting up the following proportion:

$1 atm : 760 mmHg = 0.700 atm : x$

Solving for x, we find

$x=\frac{(760 mmHg)(0.700 atm)}{1 atm}=532 mmHg$

5 0

4 years ago

Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le

kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0

3 years ago

1. What were the 3 basic concepts of government that the English brought with them to North America?

aev [14]

Answer:

1)limited system of government

2)representative system of government

3)individual rights system of government

Explanation:

the 3 basic concepts of government that the English brought with them to North America are listed in the answer

7 0

3 years ago