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2 years ago

Define mass defect and binding energy

Physics

1 answer:

natima [27]2 years ago

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Answer:

About Transcript. Nuclear binding energy is the energy required to split an atom's nucleus into protons and neutrons. Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. The binding energy of a system can appear as extra mass, which accounts for this difference.

Explanation:

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In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

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2 years ago

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lutik1710 [3]

Karma, and sensei. Maybe nagisa

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2 years ago

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You stand new the earth's equator. A positively charged particle that starts moving parallel to the surface of the earth in a st

Talja [164]

Answer:

c. from south to north

Explanation:

since we know that:

F = qv×B

where F is the force and in this case is pointed upward

where v is the velocity due east

the field must be due north by the right hand rule

Therefore, near the equator the magnetic field lines of the earth are directed north from south

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2 years ago

A 2.8-kg cart is rolling along a frictionless, horizontal track towards a 1.2-kg cart that is held initially at rest. The carts

agasfer [191]

Answer with Explanation:

We are given that

Mass of one cart, $m_1=2.8 kg$

Mass of second cart, $m_2=1.2 kg$

Initial velocity of one cart, $u_1=4.6m/s$

Initial velocity of second cart, $u_2=-2.7 m/s$

a.Total momentum, $P=m_1u_1+m_2u_2=2.8(4.6)+1.2(-2.7)$

$P=9.64 kgm/s$

b.Velocity of second cart, $v_2$ =0

According to law of conservation of momentum

Initial momentum=Final momentum

$9.64=2.8v_1+1.2\times 0$

$v_1=\frac{9.64}{2.8}$

$v_1=3.44m/s$

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2 years ago

An object's true weight is 123 N. When it is completely submerged in water, its

SashulF [63]

Object true weight is given as

$mg = 123 N$

now we know that g = 9.8 m/s^2

$m* 9.8 = 123$

$m = \frac{123}{9.8} = 12.55 kg$

now when it is complete submerged in water its apparent weight is given as 82 N

apparent weight = weight - buoyancy force

apparent weight = 82 N

weight = 123 N

now we have

82 = 123 - buoyancy force

buoyancy force = 123 - 82 = 41 N

now we also know that buoyancy force is given as

$F_b = p_{liq}Vg$

$41 = 1000*V*9.8$

$V = \frac{41}{1000*9.8}$

$V = 4.18 * 10^{-3} m^3$

now as we know that mass of the object is 12.55 kg

its volume is 4.18 * 10^-3 m^3

now we know that density will be given as mass per unit volume

$density = \frac{m}{V}$

$density = \frac{12.55}{4.18*10^{-3}$

$density = 3002.4 kg/m^3$

so here density of object is 3002.4 kg/m^3

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2 years ago

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Define mass defect and binding energy​

Define mass defect and binding energy