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lara31 [8.8K]
3 years ago
12

How does half-life apply to real life?

Physics
2 answers:
Lyrx [107]3 years ago
6 0
The term half-life is defined as the time it takes for one-half of the atoms of a radioactive material to disintegrate."
wlad13 [49]3 years ago
3 0
half-life? what do you mean
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World's biggest fandom 2020​
lesya692 [45]

Answer:

BTS was the biggest fandom in 2020

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3 years ago
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1. What is wave motion
nadya68 [22]
A wave is basically propagation of disturbances—that is, deviations from a state of rest or equilibrium—from place to place in a regular and organized way. Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties.
8 0
2 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
8_murik_8 [283]

Answer:

The gplanet is 0.193 m/s^2

Explanation:

The speed of the pulse is:

v=\frac{lengthofthewipe}{traveltime} =\frac{1.6}{0.0656} =15.24m/s

v=\sqrt{\frac{MgL}{m} } \\v^{2} =\frac{MgL}{m} \\g=\frac{mv^{2} }{ML}

where

m=mass of the wire=4 g= 4x10^-3 kg

M=mass of the object= 3 kg

Replacing values:

g=\frac{4x10^{-3}*15.24^{2}  }{3*1.6} =0.193 m/s^{2}

7 0
3 years ago
The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5t, collide with molecules of t
andreev551 [17]
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R =  m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

 

4 0
3 years ago
Net force causes motion
ASHA 777 [7]

Answer:

yes

Explanation:

8 0
2 years ago
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