Answer:
The final velocity of the thrower is 
and the final velocity of the catcher is 
.
Explanation:
Given:
The mass of the thrower, 
.
The mass of the catcher, 
.
The mass of the ball, 
.
Initial velocity of the thrower, 
Final velocity of the ball, 
Initial velocity of the catcher, 
Consider that the final velocity of the thrower is 
. From the conservation of momentum,
Consider that the final velocity of the catcher is 
. From the conservation of momentum,
Thus, the final velocity of thrower is 
and that for the catcher is 
.
According to newton's law Force = mass * acceleration
so , 100 = 50 * a
so , a= 2 m/s^2
Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
The answer is, "B", "Ammonia".
Answer:
–735.17 N
The negative sign indicate that the force is acting in opposition direction to the car.
Explanation:
The following data were obtained from the question:
Mass (m) of car = 782.10 kg
Initial velocity (u) = 7.60 m/s
Final velocity (v) = 3.61 m/s
Time (t) = 4.23 s
Force (F) =?
Next, we shall determine the acceleration of the car. This can be obtained as follow:
Initial velocity (u) = 7.60 m/s
Final velocity (v) = 3.61 m/s
Time (t) = 4.23 s
Acceleration (a) =?
a = (v – u) / t
a = (3.61 – 7.60) / 4.23
a = –3.99 / 4.23
a = –0.94 m/s²
Finally, we shall determine the force experienced by the car as shown below:
Mass (m) of car = 782.10 kg
Acceleration (a) = –0.94 m/s²
Force (F) =?
F = ma
F = 782.10 × –0.94
F = –735.17 N
The negative sign indicate that the force is acting in opposition direction to the car.
