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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<em>Resultant angle; θ = 25.59° </em>
This question is dealing with bearings and distance.
We are told that from point A, the camel walks 20 km at 15° in the south of east direction.
Thus, d_s,e = 20 km
Resolving along the horizontal east direction gives; d_e = 20 cos 15
d_e = 19.32 km
Also, resolving along the vertical south direction gives; d_s = 20 sin 15
d_s = 5.18 km
Net vertical distance; d_vert = 8km - 5.18km = 2.72 km
Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km
Now, the resultant angle is given by;
tan θ = d_vert/d_hor
tan θ = 2.72/5.68
tan θ = 0.4789
θ = tan^(-1) 0.4789
θ = 25.59°
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Answer:
Explanation:
Potential energy, which is the energy a body assumes at a position, can be calculated using the formula:
P.E = m × g × h
Where;
m = mass (kg)
g = acceleration due to gravity (10m/s²)
h = height (m)
The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
<h3>What is power of the circuit?</h3>
The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.
P = VI
Circuit A has bulbs in series while the circuit B has bulbs in parallel.
When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.
The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.
Thus, the last option is correct.
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