The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
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Answer:
Resultant force = (232.93î + 246.10j) N
x-component of the resultant force = (+232.93î) N
y-component of the resultant force = (+246.1j) N
Explanation:
The net external force on the statue is equal to the resultant force on the statue.
And the resuphant force is a vector sum of all the other forces acting on the statue.
Force 1 = (45î) N
Force 2 = (105j) N
Force 3 = (235cos 36.9°)î + (235 sin 36.9°)j = (187.93î + 141.10j) N
Resultant force = (Force 1) + (Force 2) + (Force 3)
Resultant force = 45î + 105j + (187.93î + 141.10j) = (232.93î + 246.10j) N
Hope this helps!!!
Answer: They will fall at the same time
Explanation:I had a google doc about this
Answer:
E = - 11.4 N/C, Directed Downward
Explanation:
Force = 2.00×10⁻⁵N , charge Q = –1.75μC = -1.75 ×10⁻⁶ C
To Find: Electric Field E = ? and Direction=?
Sol
we Have
E = F / Q
E = 2.00×10⁻⁵N / -1.75 ×10⁻⁶ C
E = - 11.428 N/C (-ve sign shows that it is Negative Electric Field around a negative charge and will attract the Positive Field)
As given an upward force on negative charge so direction of the Electric Filed will be Downward (as Electric field is opposite to direction of force on negative charge)
