<span>The magnitude of a is 1.5
The magnitude of b is -3
The magnitude of the vector is
</span>√(1.5² + (-3)²) = 3.35
<span>The angle is
</span>θ = tan⁻¹ (-3/1.5) = 63.43°
<span>The vector is drawn with a magnitude of 3.35 and an angle of 63.43</span>°.
Answer: the maximum heigth of the stadium at ist back wall is 151.32 ft
Explanation:
1. use the position (x) equation in parobolic movement to find the time (t)
565 ft = [frac{176 ft}{1 s\\}[/tex] * cos (35°) * t
t= 3.92 s
2. use the position (y) equation in parabolic movement to find de maximun heigth the ball reaches at 565 ft from the home plate.
y= [[frac{176 ft}{1 s\\}[/tex] * sen (35°) * 3.92 s] - 
y= 148.32 ft
3. finally add the 3 ft that exist between the home plate and the ball
148.32 ft + 3 ft = 151.32
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>
Explanation:
When the charge of 2e is placed in between the plates .
The force applied on this charge by plates is = q E
here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C
and E is the magnitude of electric field intensity
The work done = Force x displacement
Thus W = q E x S
here S is displacement
Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8
= 1.02 x 10⁻¹⁷ J
This work will be converted into the kinetic energy of charge .
Thus K.E = 1.02 x 10⁻¹⁷ J
Population density = Number of beetles / Unit Square Area
= 20 beetles / 5m² = 4 beetles per m²
