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.
Explanation:
Velocity = displacement / time
v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)
v = 0.20 m/s
Answer:
Shawn's speed relative to Susan's speed = 10 mph
Resultant velocity = 82.32 mph
Explanation:
The given data :-
i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.
ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.
iii) The speed of both Susan and Shawn is relative to earth.
iv) The angle between Susan in north and Shawn in east is 90°.
We have to find Shawn's speed relative to Susan's speed.
v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph
Resultant velocity,

v = 82.32 mph
Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N = 0.2
Hence 0.2 is the coefficient of static friction
Answer:
The final angular speed is 16.1 rad/s
Explanation:
Given;
initial moment of inertia, I₁ = 2.56 kg.m²
final moment of inertia, I₂ = 0.40 kg.m²
initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s
Apply the principle of conservation of angular momentum;
I₁ω₁ = I₂ω₂
where;
ω₂ is the final angular speed
ω₂ = (I₁ω₁) / (I₂)
ω₂ = (2.56 x 2.514) / (0.4)
ω₂ = 16.1 rad/s
Therefore, the final angular speed is 16.1 rad/s