Explanation:
Radius of a charged particle is given by
r=mv / Bq
= k/ q
where k = m v / B is a constant.
i.e. more is the magnitude of charge, less is the radius.
(inversely proportional)
From the diagram r_3 > r_2 > r_1 (more the curvature, less is the radius)
( although drawing is not given i am assuming the above order, however, one can change the order as per the diagram. The concept used remains the same)
therefore, q_1 > q_2 > q_3
.
Answer:
<em>a) Fvt cosθ</em>
<em>b) Fv cosθ</em>
<em></em>
Explanation:
Each horse exerts a force = F
the rope is inclined at an angle = θ
speed of each horse = v
a) In time t, the distance traveled d = speed x time
i.e d = v x t = vt
also, the resultant force = F cosθ
Work done W = force x distance
W = F cosθ x vt = <em>Fvt cosθ</em>
<em></em>
b) Power provided by the horse P = force x speed
P = F cosθ x v
P = <em>Fv cosθ</em>
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:
(a) 
(b) 142
(c) 
(d) 96.8 mph
(e) 0.426 s
(f) 0.061 rad
Explanation:
Velocity is a time-derivative of position.
(a) 
(b) Since is independent of 
, it follows it was constant throughout. Hence, at any point or time, the horizontal velocity is 142.
(c) 
(d) When it passes the home plate, the ball has travelled 60.5 ft (from the question). This is horizontal, so it is equivalent to 
.
.
In this time, the vertical velocity, 
is
The speed of the ball at thus point is 
ft/s
To convert this to mph, we multiply the factor 3600/5280
(e) The time has been determined from (d) above.
(f) This angle is given by
(Note here we are considering the acute angle so we ignore the negative sign)
In radians, this is
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
