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maks197457 [2]
4 years ago
7

A block of wood mass 0.60kg is balanced on top of a vertical port 2.0m high. A 10gm bullet is fired horizontally into the block

and the embedded bullet land at a 4.0m from the base of the port. Find the initial velocity of the bullet.
Physics
1 answer:
anzhelika [568]4 years ago
7 0

Answer:

Mass of bullet is m=0.01kg

Mass of the block is M=4kg

Coefficient=0.25,distance=20m

So, let the speed of the block just after the bullet embedded in it be V and v be the speed of bullet before striking the block,

By applying conservation of momentum,

mv=(m+M)V

V=

M+m

mv

Explanation:

please mark me as the brainliest answer and please follow me for more answers to your questions..

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Formula for calculating work done in machines ​
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force×distance

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Can you solve it descriptively . thanks
Solnce55 [7]

Answer:

|M_y| = 170.82 \ N.mm

Explanation:

From the diagram affixed below completes the question

Now from the diagram; We need to resolve the force at point  A into (3) components ; i.e x.y. & z directions which are equivalent to F_x \ , F_y \ ,  F_z

So;

F_x = positive x axis

F_y = Negative y axis

F_z = positive z axis

Then;

|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ?  ---- equation (3)

From equation (1); Let's make F_y the subject of the formula ; then :

F_y = \frac{77+11F_z}{27}

Substituting  the value for F_y into equation (2) ; we have:

(\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since  \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm

4 0
3 years ago
In Hooke's law, Fspring=kΔx, what does the ∆x stand for
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Answer:

Change in Displacement

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A train is 240 meters long and travels 20 m/s. How long does
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Answer:

=18 sec

Explanation:

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240÷20=12sec

if 240=12sec,find 360

360×12÷240

=18 sec

6 0
3 years ago
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