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solmaris [256]
3 years ago
10

An 8.0 g bullet is fired from a gun at 120 m/s into a 2.0 kg block of wood sitting on a fence post. If the bullet hits the wood

and remains in it, what is the speed of the block after the collision?
Physics
1 answer:
nydimaria [60]3 years ago
8 0

Answer:

v = 0.478 m/s

Explanation:

As we know that bullet strike the block and fix in it

so here if bullet and box is taken as a system then there is no external force on it

so here the initial momentum of bullet and block must be equal to final momentum of bullet and block

so here we have

m_1v_{1i} + m_2v_{2i} = m_1v + m_2v

now we have

m_1 = 8 g

v_1 = 120 m/s

m_2 = 2kg

v_{2i} = 0

now from above equation we have

(0.008)(120) + 2(0) = (0.008)v + 2v

v = 0.478 m/s

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C,

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Explain: a camel walks easily on sand surface than a man inspite of the fact that a camel is much heavier than a man
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3 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
Objects 1 and 2 attract each other with a electrostatic force of 72.0 units. If the charge of object 1 is doubled AND the charge
lbvjy [14]

Answer:

432 units

Explanation:

Let the charges be q and Q separated by a distance r. The electrostatic force , F = kqQ/r² = 72 units. If q = 2q and Q = 3Q, then the new electrostatic force is

F = k × 2q × 3Q/r² = 6kqQ/r² = 6 × 72 = 432 units

5 0
3 years ago
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