For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
 
        
                    
             
        
        
        
Answer:
Option B. 6.25 J/S 
Explanation:
Data obtained from the question include:
t (time) = 2secs 
F (force) = 50N
d (distance) = 0.25m
P (power) =? 
The power can be obtained by using the formula P = workdone/time. 
P = workdone / time
P = (50 x 0.25)/ 2
P = 6.25J/s
 
        
             
        
        
        
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s
Induced peak emf = NABω
                                = 500 x (π x 0.25²) x 0.425 x 376.738
                                = 15721.16 V
                                = 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
 
        
             
        
        
        
Answer:
The diagram represents two charges, q1 and q2, separated by a distance d. Which change would produce the greatest increase in the electrical force between the two charges? *
Explanation:
doubling charge q1, only
 
        
             
        
        
        
From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.
 We need to find its height.
We will use the formula P.E = mgh
Therefore h = P.E / mg
where P.E is the potential energy,
 m is mass in kg,  
g is acceleration due to gravity (9.8 m/s²)
 h is the height of the object's displacement in meters.
h = P.E. / mg
h = 14000 / 40 × 9.8
h = 14000 / 392
h = 35.7
Therefore the canon ball was 35.7 meters  high.