A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical component of its velocity after 4.0 s if air resistance is neglected?

Explanation:

The dart is project with 150 m/s from a point at an angle of 30⁰

The vertical component of velocity = 150 sin 30 = 75 m/s

Thus initial vertical velocity is = 75 m/s

The velocity after 4 s can be calculate by

v = u - g t

here u is the initial velocity and t is the time , g is the acceleration due to gravity .

Thus v = 75 - 10 x 4 = 35 m/s

i need the answer to this too but idk it ; -;

  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