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3 years ago

9

Using the binding energy per nucleon number that you previously calculated for the isotope cobalt-60, is this a high amount of b

inding energy per nucleon according to the graph?

1 answer:

lora16 [44]3 years ago

6 0

Answer: yes

Explanation: I did this question before

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An alpha particle (charge +2e) travels in a circular path of radius .5m in a magnetic field of 1.0 T. Find the (a) period, (b) s

navik [9.2K]

Given Information:

Radius = r = 0.5 m

Magnetic field = 1.0 T

Required Information:

Period = T = ?

Speed = v = ?

Kinetic energy = KE = ?

Answer:

Period = 0.13x10⁻⁶ seconds

speed = 24.16x10⁶ m/s

Kinetic energy = 12.11 MeV

Explanation:

(a) period

The time period of alpha particle is related to its orbital speed as

T = 2πr/v eq. 1

According to newton's law

F = ma

Force due to magnetic field is given by

F = qvB

qvB = ma

qvB = m(v²/r)

qB = mv/r

v = qBr/m eq. 2

substitute the eq. 2 in eq. 1

T = 2πr/qBr/m

r cancels out

T = 2π/qB/m

T = 2πm/qB

T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

T = 0.13x10⁻⁶ seconds

(b) speed

From equation 1

T = 2πr/v

v = 2πr/T

v = 2π*0.5/0.13x10⁻⁶

v = 24.16x10⁶ m/s

(c) kinetic energy (in electron volts)

Kinetic energy is given by

KE = 0.5mv²

KE = 0.5*6.65x10⁻²⁷*(24.16x10⁶)²

KE = 1.94x10⁻¹² J

since 1 electron volt has 1.602x10⁻¹⁹ J

KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

KE = 12.11 MeV

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3 years ago

A student wants to determine the effect of mass on kinetic energy. She will drop two balls of the same size into a pool of water

Anastasy [175]

Answer:

She should drop two balls with different masses from the same height

Explanation:

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2 years ago

I KNOW YALL SEE THIS I NEED HELP GOD WILL GIVE U MANY BLESSING HELP A POOR SOUL OUT THE RECENT QUESTIONSSSSSSSSSSSSS. ://///

photoshop1234 [79]

Answer:

ok

Explanation:

will do

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2 years ago

Read 2 more answers

A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Which number marks a crest?

lyudmila [28]

number 2 because the curve demstrates the crest GOOD LUCK i hope i got you the correct answer if not sorry

5 0

3 years ago