Answer: 
beta particles
Explanation:
Given mass = 14.0 g
Molar mass = 137 g/mol
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 
of particles.
1 mole of cesium contains atoms =
0.102 moles of cesium contains atoms = 
The relation of atoms with time for radioactivbe decay is:
Where 
=atoms left undecayed
= initial atoms
t = time taken for decay = 3 minutes
= half life = 30.0 years = 
minutes
The fraction that decays : 
Amount of particles that decay is = 
Thus beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.
Answer:
a. 7.046 Nm²/C
b. 2.348 Nm²/C
Explanation:
Data given:
Base of equilateral triangle = 25.0 cm = 0.25 m
Strength of electric field = 260 N/C
In order to find the electric flux we first have to find out the area of triangle.
Area of triangle = 
= 
= 0.0271 m³
Lets find electric flux,
Electric Flux = E. A
= 260×0.0271
= 7.046 Nm²/C
Now we can find the electric flux through each of the three sides.
Electric flux through three sides = 
= 2.348 N m²/C
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