(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>
Horizontal and vertical components of the ball's velocity</h3>
Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>

Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>
Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

<h3>Resultant speed of the ball and crosswind</h3>

<h3>Distance off course the ball would be carried</h3>
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671
Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
Answer:
E = 420.9 N/C
Explanation:
According to the given condition:

where,
E = Magnitude of Electric Field = ?
v = speed of charge = 230 m/s
B = Magnitude of Magnetic Field = 0.61 T
θ = Angle between speed and magnetic field = 90°
Therefore,

<u>E = 420.9 N/C</u>