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rhe12345678913p50uea
31.07.2020 •
Physics
A 20 g bullet strikes a 1.5 kg wood block and becomes embedded in it (all horizontal motion). The coefficient of friction between the block and the surface is 0.3 and the impact makes the block move a distance of 7 m before it comes to rest. What was the muzzle speed (in m/s) of the bullet? Report your answer with two significant figures.
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Ответ:
Explanation:
Kinetic frictional force acting on the block = μ mg
where μ is coefficient of friction , m is mass of block.
.3 x 1.5 x 9.8 = 4.41 N .
Let v be the velocity of bullet + block after collision
kinetic energy of composite mass after the strike
= 1 /2 x 1.52 x v²
this will be equal to work done by friction .
.76 v² = 4.41 x 7
v² = 40.62
v = 6.37 m /s
Now we can obtain muzzle speed of bullet by applying conservation of momentum .
Let this speed be u
initial momentum of bullet
= .02 x u
final momentum of composite mass
= 1.52 x 6.37
.02 u = 1.52 x 6.37
u = 484.12 m /s .
= 480 m /s ( in two significant figures )
Ответ:
All the cell come from preexisting cells.
Organism can have one or more cells
Explanation: