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svetoff [14.1K]
3 years ago
15

A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, αα-4. The numbers indicate the

inertias of the atoms and the alpha particle in atomic mass units (1 amuamu = 1.66 × 10−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.2 × 10^5 m/s.
How much of the uranium atom's internal energy is released in the breakup?
Physics
1 answer:
alexdok [17]3 years ago
3 0

Answer: E = 5.80*10^-13 J

Explanation:

Given

We use the law of conservation of momentum to solve this

Momentum before breakup = momentum after breakup

0 = m1v1 + m2v2

0 = 238m * -2.2*10^5 + 4m * v2

0 = -523.6m m/s + 4m * v2

v2 * 4m = 523.6m m/s

v2 = 523.6 m m/s / 4m

v2 = 130.9*10^5 m/s

v2 = 1.31*10^7 m/s

Using this speed in the energy equation, we have

E = 1/2m1v1² + 1/2m2v2²

E = 1/2 * (238 * 1.66*10^-27) * -2.2*10^5² + 1/2 * (4 * 1.66*10^-27) * 1.31*10^7²

E = [1/2 * 3.95*10^-25 * 4.84*10^10] + [1/2 * 6.64*10^-27 * 1.716*10^14]

E = (1/2 * 1.911*10^-14) + (1/2 * 1.139*10^-12)

E = 9.56*10^-15 + 5.7*10^-13

E = 5.80*10^-13 J

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A bucket that has a mass of 20 kg when filled with sand needs to be lifted to the top of a 15 meter tall building. You have a ro
prohojiy [21]

Answer:

work done lifting the bucket (sand and rope) to the top of the building,

W=67.46 Nm

Explanation:

in this question we have given

mass of bucket=20kg

mass of rope=.2\frac{kg}{m}

height of building= 15 meter

We have to find the work done lifting the bucket (sand and rope) to the building =work done in lifting the rope + work done in lifting the sand

work done in lifting the rope is given as,W_{1}=Force \times displacement

=\int\limits^{15}_0 {.2x} \, dx ..............(1)

=.1\times 15^2

=22.5 Nm

work done in lifting the sand is given as,W_{2}=Force \times displacement

W_{2}=\int\limits^{15}_0 F \, dx.................(2)

Here,

F=mx+c

here,

c=20-18

c=2

m=\frac{20-18}{15-0}

m=.133

Therefore,

F=.133x+2

Put value of F in equation 2

W_{2}=\int\limits^{15}_0 (.133x+2) \, dx

W_{2}=.133 \times 112.5+2\times15\\W_{2}=14.96+30\\W_{2}=44.96 Nm

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work done lifting the bucket (sand and rope) to the top of the building,W=W_{1}+W_{2}

W=22.5 Nm+44.96 Nm

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4 0
3 years ago
A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate. The plates have a
butalik [34]

Answer:

The potential difference between the plates is 596.2 volts.

Explanation:

Given that,

Capacitance C=260\ pF

Charge q=0.155\ \mu\ C

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We need to calculate the potential difference between the plates

Using formula of potential difference

V= \dfrac{Q}{C}

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C = capacitance

Put the value into the formula

V=\dfrac{0.155\times10^{-6}}{260\times10^{-12}}

V=596.2\ volts

Hence,The potential difference between the plates is 596.2 volts.

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