If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy
<span>v1 = velocity of #1 </span>
<span>v1' = velocity of #1 after collision </span>
<span>v2' = velocity of #2 after collision. </span>
<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
<span>5^2 = 4.35^2 + v2' ^2 </span>
<span>v2 = 2.46 m/s <--- ANSWER</span>
Answer:
the total momentum is 8 .2 kg m/s in north direction.
Explanation:
given,
mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s
mass(m₂) 4.00 kg, moving south at v₂ = 3.70 m/s
mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s
north as the positive axis
south as the negative axis
now
total momentum = m₁v₁ + m₂ v₂ + m₃ v₃
total momentum = 3 x 3 - 4 x 3.7 + 7 x 2
= 9 - 14.8 + 14
= 8 .2 kg m/s
hence, the total momentum is 8 .2 kg m/s in north direction.
Answer:
12 units
Explanation:
This problem can be solved if we take into account the equation for a sphere

where we took that the radius is 13 units. If we take z=5 and we replace this value in the equation of the sphere we have

where we have taken x2 +y2 because if the equation of a circunference.
In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.
Hence, the radius is 12 units
I hope this is useful for you
regards
I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.
Work = 20 N * 6 m * 1
Work = 120 Nm
Work = 120 joules
Hope this helps!