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3 years ago

What is step one of a star called?

1 answer:

3 0

Answer: Stage 1- Stars are born in a region of high density Nebula, and condenses into a huge globule of gas and dust and contracts under its own gravity. This image shows the Orion Nebula or M42 . Stage 2 - A region of condensing matter will begin to heat up and start to glow forming Protostars.

Explanation:

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Elena (60.0 kg) and Madison (65.0 kg) are ice-skating at the Rockefeller ice rink in New Yok city. Their friend Tanner sees Elen

Bas_tet [7]

1. +72.0 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

Taking "to the right" as positive direction, for Elena we have

m = 60.0 kg is the mass

v = +1.20 m/s is the velocity

So, Elena's momentum is

$p_e=(60.0 kg)(+1.20 m/s)=+72.0 kg m/s$

2. -162.5 kg m/s

Here Madison is moving in the opposite direction of Elena (to the left), so her velocity is

v = -2.50 m/s

while her mass is

m = 65.0 kg

Therefore, her momentum is

$p_m= (65.0 kg)(-2.50 m/s)=-162.5 kg m/s$

3. -90.5 kg m/s

The total momentum of Elena and Madison is equal to the algebraic sum of their momenta; taking into account the correct signs, we have:

$p=p_e + p_m = +72.0 kg m/s - 162.5 kg m/s =-90.5 kg m/s$

4. 0.72 m/s to the left

We can find the final speed of Elena and Madison by using the law of conservation of momentum. In fact, the final momentum must be equal to the initial momentum (before the collision).

The initial momentum is the one calculated at the previous step:

$p_i = -90.5 kg m/s$

while the final momentum (after the collision) is given by

$p_f = (m_e + m_m) v$

where

$m_e$ is Elena's mass

$m_m$ is Madison's mass

v is their final velocity

According to the law of conservation of momentum,

$p_i = p_f\\p_i = (m_e + m_m) v$

So we can find v:

$v=\frac{p_i}{m_e + m_m}=\frac{-90.5 kg m/s}{60.0 kg+65.0 kg}=-0.72 m/s$

and the direction is to the left, since the sign is negative.

8 0

3 years ago

A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he

juin [17]

The skater's final angular speed is equal to 12 rad/s.

When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.

The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.

In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.

Learn more about angular momentum here brainly.com/question/7538238

#SPJ4

7 0

1 year ago

The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.

Otrada [13]

Answer:

a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

I = ∫ F dt = Δp

F t = m $v_{f}$ - m v₀

F = m (v_{f} -v₀o) / t

in the exercise indicates that the speed module is the same, but in the opposite direction

F = m (-2v) / t

if we use Newton's second law

F = m a

we substitute

- 2 mv / t = m a

a = - 2 v / t

let's calculate

a = - 2 4.59 10²/3 10⁻¹³

a = 3.06 10¹⁵ m / s

b) F= m a

F= 4.68 10⁻²⁶ 3.06 10¹⁵

F= 1.43 10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

F_total = n F

F_total = n m a

F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵

F_total = 14.32 10⁻²⁶ N

8 0

3 years ago

Suppose a ball had a potential energy of 5 J when you dropped it.What would be it’s kinetic energy just as it hit the ground?

timama [110]

Answer:

Due to Conservation of Energy just as the ball hits the ground it's potential energy is assumed zero

Therefore disregarding air resistance all energy is converted into potential energy.

So KE = PE

(5 J)

4 0

3 years ago

A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given t

Mademuasel [1]

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

$= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV$

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

8 0

2 years ago

What is step one of a star called?​

What is step one of a star called?