Answer:
They are both wrong!

Liquid oxygen really is a pale blue color.
I’ve seen it.

And they cant say that liquid and solid oxygen is blue which makes the sky blue because they’re not and it doesn’t make up for the color of the sky.

The sky is actually blue because It reflects more light than It can absorb
aka Rayleigh Scattering.

-HOPE THAT HELPED

8 0

3 years ago

a rifle is fired and recoils when the bullet leaves the gun. this is an example of newton's 3rd law. the force on the bullet is.

Alex_Xolod [135]

The force of the bullet is the same.

3 0

3 years ago

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A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the

Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$ ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$ (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

7 0

3 years ago