Adaring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. the constant speed of the horse is 14.5 m/s, and the distance from the limb to the level of the saddle is 3.29 m.
(a) what must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) how long is he in the air?

For Part A:

ΔX=11.8813 m

Part B:

Explanation:

Note: In order to find Part A we first have to find Part B i.e time

Data given:

a=-9.8m/s^2 (-ve is because ranch is falling down)

Δy=-3.29m (-ve is because ranch is falling down)

Second equation of Motion:

Δy=

V_i=0, Equation will become

Δy=

For Part A:

Again:

Second equation of Motion:

ΔX=

Since velocity is constant a=0

V_i=14.5m/s, t=0.8194 sec

ΔX=

ΔX=

ΔX=11.8813 m

Part B: (Calculated above)

Data given:

a=-9.8m/s^2 (-ve is because ranch is falling down)

Δy=-3.29m (-ve is because ranch is falling down)

Second equation of Motion:

Δy=

V_i=0, Equation will become

Δy=

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

spare some change and give it to the poor

explanation:

plz : ((