A wagon is pulled at an angle of 30 degrees to the horizontal.
You need to have the Mass and velocity
The pyrite will be bigger, because its density is much lower.
I <em>do</em> know that the gold's volume will be 2.5906 (With a bunch more numbers after it)
50 divided by 19.3 = 2.5906
Answer:
a.) L = 2.64 kgm^2/s
b.) V = 4.4 m/s
Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.
So,
Radius r = 0.6 m
Mass M = 2 kg
Velocity V = 1.1 m/s
Angular momentum L can be expressed as;
L = MVr
Substitute all the parameters into the formula
L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1
the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1
b. If she pulls her arms into 0.15 m,
New radius = 0.15 m
Using the same formula again
L = 2( MVr)
2.64 = 2( 2 × V × 0.15 )
1.32 = 0.3 V
V = 1.32/0.3
V = 4.4 m/s
Her new linear speed will be 4.4 m/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)]