2.71 m/s fast Hans is moving after the collision.
<u>Explanation</u>:
Given that,
Mass of Jeremy is 120 kg (
)
Speed of Jeremy is 3 m/s (
)
Speed of Jeremy after collision is (
) -2.5 m/s
Mass of Hans is 140 kg (
)
Speed of Hans is -2 m/s (
)
Speed of Hans after collision is (
)
Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is
= 
Substitute the given values,
= 120 × 3 + 140 × (-2)
= 360 + (-280)
= 80 kg m/s
Linear momentum after the collision of Jeremy and Hans is
= 
= 120 × (-2.5) + 140 × 
= -300 + 140 × 
We know that conservation of liner momentum,
Linear momentum before the collision = Linear momentum after the collision
80 = -300 + 140 × 
80 + 300 = 140 × 
380 = 140 × 
380/140= 
= 2.71 m/s
2.71 m/s fast Hans is moving after the collision.
Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
</span>
Hope this helps :))
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Explanation:
answer 689910
M x G x H
mass = 77
G= 9.8
Meter = 915 this is the height
77 x 9.8= 754.6
754.6 X 915 = 689910 J
Answer:

Explanation:
Given data
Speed of jet Vjet=1190 km/h
Speed of prop driven Vprop=595 km/h
Height of jet 7.5 km
Height of prop driven transport 3.8 km
Density of Air at height 10 km p7.8=0.53 kg/m³
Density of air at height 3.8 km p3.8=0.74 kg/m³
The drag force is given by:

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:

Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points