The kinematic equations of motion that apply here are<span>y(t)=votsin(θ)−12gt2</span>and<span>x(t)=votcos(θ)</span>Setting y(t)=0 yields <span>0=votsin(θ)−12gt2</span>. If we solve for t, we obtain, by factoring,<span>t=<span>2vsin(θ)g</span></span>Substitute this into our equation for x(t). This yields<span>x(t)=<span><span>2v2cos(θ)sin(θ)</span>g</span></span><span>This is equal to x=<span><span>v^2sin(2θ)</span>g</span></span>Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. <span> 60.23°, 29.77° </span>
Answer:
The work done against gravity is 78.4 J
Explanation:
The work is calculated by multiplying the force by the distance that the
object moves
W = F × d, where W is the work , F is the force and d is the distance
The SI unit of work is the joule (J)
We need to find the work done against gravity when lowering a
16 kg box 0.50 m
→ F = mg
→ m = 16 kg, and g = 9.8 m/s²
Substitute these value in the rule
→ F = 16 × 9.8 = 156.8 N
→ W = F × d
→ F = 156.8 N and d = 0.50
Substitute these values in the rule
→ W = 78.4 J
<em>The work done against gravity is 78.4 J</em>
Answer:
Carbon dioxide and water
Explanation:
The products of complete combustion are always carbon dioxide and water.
The balanced reaction is:
4 CH₃OH + 3 O₂ → 4 CO₂ + 2 H₂O
The answer would be D hope it helps and sorry if it is wrong. :)
Answer:
B
Explanation:
V=IR I= curren V=volt R=resistor
8=2.R 8/2=R R=4
