Answer:
a. a=33.34ms⁻², V=164.4m/s
Explanation:
Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.
Below are the data given
Distance, s=404.5m,
time taken,t=4.922secs
Using the equation
S=ut+1/2at²
where u is the initial velocity and u=0
Making the acceleration the subject of the formula, we arrive at
a=2s/t²
a=(2*404.5)/4.922²
a=33.34ms⁻².
To determine the velocity, we use
V=u+at
V=0+33.34ms⁻² *4.922sec
V=164.4m/s
Metallic bonding<span> is the force of attraction between valence electrons and the metal ions. It is the sharing of many detached electrons between many positive ions,
Hopefully this can help you understand
</span>
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
Answer:
Upward
Explanation:
We are given that
Mass of scarp paper,
1mg=
Distance,d =8 mm=
Magnitude of electric force =
Where 
Substitute the values
Gravitational force act in downward direction.
The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.
Hence, the direction of electric force is upward.
<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>
