The answer is c. it requires no works
I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.
Answer:
Explanation:
Gravitational force between two objects having mass m₁ and m₂ at a distance R
F = G m₁ m₂ / R²
Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²
= 1.01 x 10⁻⁶ N
b )
Force between baby and Jupiter
F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²
= 1.31 x 10⁻⁶ N
c )
Ratio = 1.01 / 1.31
= .77
Answer:
B) x^2+6x+8
Explanation:
x-4 | x^3+2x^2-16x-32
- x^3-4x^2 <-- (x-4)(x^2)
_________________
6x^2-16x-32
- 6x^2-24x <-- (x-4)(6x)
_________________
8x-32
- 8x-32 <- (x-4)(8)
___________________________
0 | x^2+6x+8
This means the answer is B) x^2+6x+8
<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>