Answer:
See the answers below.
Explanation:
Momentum is defined as the product of mass by velocity, and can be calculated by means of the following expression.
where:
P = Momentum [kg*m/s]
m = mass = 10 [g] = 0.01 [kg]
v = velocity = 400 [m/s]
i)
![P=0.01*400\\P=4[kg*m/s]](https://tex.z-dn.net/?f=P%3D0.01%2A400%5C%5CP%3D4%5Bkg%2Am%2Fs%5D)
ii)
The momentum of the gun is equal to zero, because it does not move before being fired, the weapon only moves after having fired the weapon.
iii)
Since the momentum is conserved before and after the shot, the same momentum given to the bullet is equal to the momentum received by the gun.
![v_{recoil}=P/m\\v_{recoil}= 4/2\\v_{recoil}=2[m/s]](https://tex.z-dn.net/?f=v_%7Brecoil%7D%3DP%2Fm%5C%5Cv_%7Brecoil%7D%3D%204%2F2%5C%5Cv_%7Brecoil%7D%3D2%5Bm%2Fs%5D)
Answer:
the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms
Explanation:
Given the data in the question;
first we determine the rotational latency
Rotational latency = 60/(3600×2) = 0.008333 s = 8.33 ms
To get the longest time, lets assume the sector will be found at the last track.
hence we will access all the track, meaning that 127 transitions will be done;
so the track changing time = 127 × 15 = 1905 ms
also, we will look for the sectors, for every track rotations that will be done;
128 × 8.33 = 1066.24 ms
∴The Total Time = 1066.24 ms + 1905 ms
Total Time = 2971.24 ms
Therefore, the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms
Explanation:
Given that,
A student covered a distance of 210 meters in 35 seconds.
We need to find the student's speed in meters/second and also in meters/minute.
Speed, v = distance (d)/time (t)
So,
We know that, 1 minute = 60 seconds
Hence, the student's speed is 6 m/s or 360 meters/minute.
To the right, if you push something left the friction is to the right
It's a graph of a vehicle's fuel efficiency ... what operating speeds make it use more or less gas to go 1 mile.
