a. The particle has position vector


b. Its velocity vector is equal to the derivative of its position vector:

c. At
, the particle has position


That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of
away from the origin in a direction of
relative to the positive
axis.
d. The speed of the particle at
is the magnitude of the velocity at this time:


Then its speed at this time is

Answer:
what is that supposed to even mean
Explanation:
Answer:
A three valve manifold serves to protect the capsule from being exposed to pressures above the acceptable range and also enables transmitter isolation from the loop of the process
In closed tank level measurement installations the pressure of the gas if presence of gas on top of the liquid will need to be taken account by the use of a reference connection from the level transmitter to the tank top. The valve used in pressure equalization is kept close when the process system is operating normally.
In the event that the transmitter is removed or put into service, it is necessary to ensure that the gas phase which is at the higher pressure is does not over-range the Differential Pressure DP capsule
Explanation:
Answer: 86.47 g of carbon-14 must have been present in the sample 11,430 years ago.
Explanation:
Half-life of sample of carbon -14= 5,730 days

Let the sample present 11,430 years(t) ago = 
Sample left till today ,N= 0.060 g

![ln[N]=ln[N]_o-\lambda t](https://tex.z-dn.net/?f=ln%5BN%5D%3Dln%5BN%5D_o-%5Clambda%20t)
![\log[0.060 g]=\log[N_o]-2.303\times 0.00012 day^{-1}\times 11,430 days](https://tex.z-dn.net/?f=%5Clog%5B0.060%20g%5D%3D%5Clog%5BN_o%5D-2.303%5Ctimes%200.00012%20day%5E%7B-1%7D%5Ctimes%2011%2C430%20days)
![\log[N_o]=1.9369](https://tex.z-dn.net/?f=%5Clog%5BN_o%5D%3D1.9369)

86.47 g of carbon-14 must have been present in the sample 11,430 years ago.
Answer:
Vf = 41.6 [m/s]
Explanation:
To solve this problem we must use the equations of kinematics.
Vf² = Vo² + (2*g*y)
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
g = gravity acceleration = 9.81 [m/s²]
y = height = 88.2 [m]
Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.
Vf² = Vo² + (2*g*y)
Vf² = 0 + (2*9.81*88.2)
Vf = (1730.48)^0.5
Vf = 41.6 [m/s]