A 5 kg fish swimming at 1 m/s swallows an absentminded 500 g fish swimming toward it at a velocity that brings both fish to a ha
1 answer:
To solve this problem we will apply the concepts related to the conservation of momentum. Momentum is defined as the product between mass and velocity of each body. And its conservation as the equality between the initial and final momentum. Mathematically described as
Here
= Mass of big fish
= Mass of small fish
= Velocity of big fish
= Velocity of small fish
= Final Velocity
The big fish eats small fish and the final velocity is zero. Rearrange the equation for the initial velocity of small fish we have
Replacing we have,
The negative sign indicates that the small fish is swimming in the direction opposite to that of the big fish.
Therefore the speed of the small fish is 10m/s
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Answer:
Therefore the surface area of the balloon is increased at 4 cm³/s.
Explanation:
The balloon is being filled with air at a rate of 10 cm³/s
It means the volume of the balloon is increased at a rate 10 cm³/s.
i.e
Consider r be the radius of the balloon.
The volume of of a sphere is
Differentiate with respect to t
The surface of area of the balloon is(S) =
Differentiate with respect to t
Putting the value of
Given that r = 5 cm
=4 cm³/s
Therefore the surface area of the balloon is increased at 4 cm³/s.
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)]