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The rank is : ![P_B>P_A=P_C](https://tex.z-dn.net/?f=P_B%3EP_A%3DP_C)
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We need to rank the objects according to the magnitude of their momentum.
We know momentum ,
{ here m is mass and v is velocity }
Momentum of object A , ![P_A = m v.](https://tex.z-dn.net/?f=P_A%20%3D%20m%20v.)
Momentum of object B , ![P_B= \dfrac{m}{2}\times4v=2\ mv.](https://tex.z-dn.net/?f=P_B%3D%20%5Cdfrac%7Bm%7D%7B2%7D%5Ctimes4v%3D2%5C%20mv.)
Momentum of object C , ![P_C= 3m\times\dfrac{v}{3}=mv.](https://tex.z-dn.net/?f=P_C%3D%203m%5Ctimes%5Cdfrac%7Bv%7D%7B3%7D%3Dmv.)
Now, we can rank them in order of their magnitude of momentum .
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Hence, this is the required solution.
Learn More :
Momentum
brainly.com/question/7957458
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is
![v_i=180~m/s](https://tex.z-dn.net/?f=v_i%3D180~m%2Fs)
, the final velocity is
![v_i=0~m/s](https://tex.z-dn.net/?f=v_i%3D0~m%2Fs)
, and the total time of the motion is
![\Delta t=0.02~s](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.02~s)
, so the acceleration is given by
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
![S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m](https://tex.z-dn.net/?f=S%3Dv_i%20t%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2%20%3D%20180~m%2Fs%20%5Ccdot%200.02~s%20%2B%20%5Cfrac%7B1%7D%7B2%7D%28-9000~m%2Fs%5E2%29%280.02~s%29%5E2%3D1.8~m)
So, the bullet penetrates the sandbag 1.8 meters.
Answer:
Earth's crust, called the lithosphere, consists of 15 to 20 moving tectonic plates. The heat from radioactive processes within the planet's interior causes the plates to move, sometimes toward and sometimes away from each other. This movement is called plate motion, or tectonic shift.
beainliest?
Answer:
a = -4/5 m/s^2
Explanation:
Acceleration = change in velocity / time
change in velocity = final velocity - initial velocity
a = (20 m/s - 60 m/s) / 50 s
a = -40 m/s / 50 s
a = -4/5 m/s^2
hope this helps! <3
Answer:
the moment of inertia with the arms extended is Io and when the arms are lowered the moment
I₀/I > 1 ⇒ w > w₀
Explanation:
The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,
L₀ = L_f
I₀ w₀ = I w
w =
w₀
where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms
I₀ = I_cylinder + 2 m r²
where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.
If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be
I <I₀
I₀/I > 1 ⇒ w > w₀
therefore the angular velocity (rotations) must increase
in this way the skater can adjust his spin speed to the musician.