Answer:
A) a = 2.31[m/s^2]; B) t = 14.4 [s]
Explanation:
We can solve this problem using the kinematic equations, but firts we must identify the data:
Vf= final velocity = take off velocity = 120[km/h]
Vi= initial velocity = 0, because the plane starts to move from the rest.
dx= distance to run = 240 [m]
![v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bi%7D%20%5E%7B2%7D%2B2%2Ag%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%3D120%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A%5Cfrac%7B1hr%7D%7B3600sg%7D%20%2A%20%5Cfrac%7B1000m%7D%7B1km%7D%20%3D33.33%5Bm%2Fs%5D%5C%5C%5C%5CReplacing%5C%5C33.33%5E%7B2%7D%3D0%2B2%2Aa%2A%28240%29%5C%5C%20a%3D%5Cfrac%7B11108.88%7D%7B2%2A240%7D%5C%5C%20%20a%3D2.31%5Bm%2Fs%5E2%5D%5C%5C)
To find the time we must use another kinematic equation.
![v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Creplacing%3A%5C%5C33.33%3D0%2B%282.31%2At%29%5C%5Ct%3D%5Cfrac%7B33.33%7D%7B2.31%7D%5C%5C%20t%3D14.4%5Bs%5D)
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s
Answer:
Net force on the block is 32 N.
Acceleration of the object is 6.4 m/s².
Explanation:
Let the acceleration of the object be
m/s².
Given:
Mass of the block is, 
Force of pull is, 
Frictional force on the block is, 
The free body diagram of the object is shown below.
From the figure, the net force in the forward direction is given as:

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².
Explanation:
First we will convert the given mass from lb to kg as follows.
157 lb = 
= 71.215 kg
Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

= 12818.7 mg
Convert the % of (w/w) into % (w/v) as follows.
0.65% (w/w) = 
= 
= 
Therefore, calculate the volume which contains the amount of caffeine as follows.
12818.7 mg = 12.8187 g = 
= 1972 ml
Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.
Answer:

Explanation:
We could use the following suvat equation:

where
s is the vertical displacement of the coin
v is its final velocity, when it hits the water
t is the time
g is the acceleration of gravity
Taking upward as positive direction, in this problem we have:
s = -1.2 m

And the coin reaches the water when
t = 1.3 s
Substituting these data, we can find v:

where the negative sign means the direction is downward.