Answer: 90 m/s
Explanation:
Given
mass of racecar 
velocity of racecar 
mass of still honeybadger 
after collision race car is traveling at a speed of 
conserving linear momentum
![Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]](https://tex.z-dn.net/?f=Mu%2Bm%5Ctimes0%3DMv_1%2B%20mv_2%5Cquad%5Bv_2%3D%5Ctext%7Bvelocity%20of%20honeybadger%20after%20colllision%7D%5D)


Answer:
The work done on the canister by the 5.0 N force during this time is
54.06 Joules.
Explanation:
Let the initial kinetic energy of the canister be
KE₁ =
=
= 19.44 J in the x direction
Let the the final kinetic energy of the canister be
KE₂ =
=
= 73.5 J in the y direction
Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final
= 73.5 J - 19.44 J = 54.06 J
Volumetric cylinders and volumetric flasks
Answer:
wave length is 1.2m
Explanation:
since formula of wave length is v/f
v(speed of sound in air at stp is 300ms^-1)
f(frequency 250hertz)
then wave length is 300÷250 which give 1.2m
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.