Based on your problem where as ask for the distance of the ball drop between the pitchers mound and the home plate and with a given of the speed of ball is 43m/s and the homeplates is 60.6ft away. Based on my step by step procedure and also considering the value of gravity by 9.8m/s^2 i came up with the distance of 144m away
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:
t = 7,8 s
Explanation:
From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .
When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit
Then for the cat arrives at 3,9 m/s nedds
v = vo + a*t vo = 0 then v = a*t
3,9 ( m/s) = 0,5 ( m/s² ) * t
t = 7,8 s
v = 3,9 m/s =
Answer:
Magnitude of the force on proton = F = 1.1085 × 10^-15 N
Explanation:
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
We know that,
F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)
F = 1.1085 × 10^-15 N