Answer: A) 2 B) 4 C) 1
Explanation:
The electric field associated with a parallel-plate capacitor is defined as:
A) E=Q/(L^2 * ε0); if we double the charge, the resultant electric field becomes twice as strong as its initial value.
B) Referring to the earlier electric field expression, if the plate size is doubled, the final electric field will be a quarter of the original strength.
C) If the separation distance between the plates is increased twofold, the resulting electric field remains unchanged from its initial state.
<span>Let F represent the maximum thrust produced by the car's motor. Thus, F = ma = 1300 x 3.0 = 3900 N. After adding the load, F stays the same, leading to the equation F = 1700a, which results in a = F/1700 = 3900/1700 = 2.3 m/s².</span>
Answer:
θ = 61.3°
Alicia must swim at an angle of 61.3°
Explanation:
Parameters given include:
Width of the river = 100 m
Alicia's speed in still water = 2.5 m/s
Speed of river's current = 1.2 m/s
The angle she needs to swim can be determined by combining the velocities, taking into account the current's influence.
Her swimming speed aimed against the current must offset the current's velocity;
2.5cosθ - 1.2 = 0
2.5cosθ = 1.2
cosθ = 1.2/2.5
θ = cosinverse(1.2/2.5)
θ = 61.3°
The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height.
<span>In this scenario, you can demonstrate it as follows: </span>
<span>the initial velocity is Vo </span>
<span>the launch angle is α </span>
<span>the initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity becomes </span>
<span>Vh = Vo×cos(α) </span>
<span>the total flight duration is the period required to return to a height of 0 m, thus </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 = gravitational acceleration = g (= -9.8 m/s²) </span>
<span>therefore </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 let's examine the horizontal distance. </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>therefore </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 points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set 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>As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
