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daser333 [38]
3 years ago
5

Two forces and are applied to an object whose mass is 9.31 kg. The larger force is . When both forces point due east, the object

's acceleration has a magnitude of 0.626 m/s2. However, when points due east and points due west, the acceleration is 0.340 m/s2, due east. Find (a) the magnitude of and (b) the magnitude of . (a) Number Entry field with correct answer 4.5 Units Entry field with correct answer
Physics
1 answer:
kondor19780726 [428]3 years ago
5 0

Answer:

the magnitudes of the two forces are 45.N and 1.5 N

Explanation:

This in a problem of composition of forces, as the forces are on a single axis, East-West, we can work as a problem of a single dimension, write Newton's second law for each case n

   • Two forces to the east

        F1 + F2 = m a1

  • One force to the east and the other to the west

        F1 - F2 = m a2

We see that we have a system of two equations with two unknowns, we proceed to solve it

We add sides equations

        F1 + F2 = m a1

        F1 - F2 = m a2

       2F1 = m (a1 + a2)

       F1 = m (a1 + a2) / 2

       F1 = 9.31 (0.626 +0.340) / 2

       F1 = 4.5 N

We substitute in one of the two equations and find the other force

       F1 + F2 = m a1

       F2 = m a1 - F1

       F2 = 9.31 0.646 - 4.5

       F2 = 1.5 N

     These are the magnitudes of the two forces 45.N and 1.5 N

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Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart.
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Answer:

1.6\times 10^{-7} N

2.4\times 10^{-7} N

Explanation:

i_{1} = 1 A

i_{2} = 4 A

r = distance between the two wire = 5 m

F = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}

F = (10^{-7})\frac{2(1)(4)}{5}

F = 1.6\times 10^{-7} N

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Magnetic field midway between the two wires is given as

B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))

B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))

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A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti
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Answer:

<u>\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}</u>

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Static frictional force = 68.6 N is less than 100 N applied  

10 kg block will slide on 40 kg slab and net force on it  

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=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)

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10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}

10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}

10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}

\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }

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40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}

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\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}

3 0
3 years ago
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