Answer:
Explanation:
ACCORDING TO NEWTONS SECOND LAW;
F = mass * acceleration
F = m(v-u/t)
m is the mass = 0.15kg
v is the final velocity = 11m/s
u is the initial velocity = 0m/s
t is the time = 0.015
Substitute;
F = 0.15(11-0)/0.015
F = 0.15(11)/0.015
F = 1.65/0.015
F = 110N
Hence the net force is 110N
If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.
The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.
- A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m
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Answer:0kgm/s
Explanation:
Momentum before collision=momentum after collision
Since the momentum of the two blocks have positive sign, it means they are moving in thesame direction
Therefore we use the formula
Momentum (A)+momentum (B)=Momentum (A)+momentum (B)
25+35=60+momentum (B)
60=60+momentum (B)
Momentum (B)=60-60
Momentum (B)=0kgm/s
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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The answer would be letter choice B
