<span>A) x = 41t
The classic equation for distance is velocity multiplied by time. And unfortunately, all of your available options have the form of that equation. In fact, the only difference between any of the equations is what looks to be velocity. And in order to solve the problem initially, you need to divide the velocity vector into a vertical velocity vector and a horizontal velocity vector. And the horizontal velocity vector is simply the cosine of the angle multiplied by the total velocity. So
H = 120*cos(70) = 120*0.34202 = 41.04242
So the horizontal velocity is about 41 m/s. Looking at the available options, only "A" even comes close.</span>
Answer:
Part a)

Part b)

Part c)

Part d)
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Explanation:
Part a)
When elevator is ascending with constant speed then we will have



So it will read same as that of the mass

Part b)
When elevator is decending with constant speed then we will have



So it will read same as that of the mass

Part c)
When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have



Reading is given as



Part d)
Here the speed of the elevator is constant initially
from t = 0 to t = 4.9 s
so the reading of the scale will be same as that of weight of the block
Then its speed will reduce to zero in next 3.2 s
from t = 4.9 to t = 8.1 s
The reading of the scale will be less than the actual mass
Answer:
The answer is d i think so yea
Explanation:
hope u hae a really good day
Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,

The expression of
will be,

Now put all the values of partial pressure, we get


Therefore, the partial pressure of
is, 67.009 atm
Answer:
28.15
Explanation:
the ateps is in the photo