Answer:
t=37 mins -> 2220sec
We want "T" which is the pendulum time constant
Using this equation
.5A=Ae^(-t/T)
The .5A is half the amplitude
Take ln of both sides to get ride of Ae
=ln(.5)=-2220/T
Now rearrange to = T
T=-2220/ln(.5) = 3202.78sec / 60 secs = 53.38 mins -> first part of the answer.
The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!
Answer:
0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.
Explanation:
First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:
P₁ = F/A
where,
P₁ = Pressure exerted by empty rack = ?
F = Force exerted by empty rack = Weight of Empty Rack = 40 lb
A = Base Area = 452.4 in²
Therefore,
P₁ = 40 lb/452.4 in²
P₁ = 0.088 psi
Now, we calculate the pressure exerted by the rack along with the coat.
P₂ = F/A
where,
P₂ = Pressure exerted by rack filled with coats= ?
F = Force exerted by filled rack = Weight of Filled Rack = 65 lb
A = Base Area = 452.4 in²
Therefore,
P₂ = 65 lb/452.4 in²
P₂ = 0.144 psi
Now, the difference between both pressures is:
ΔP = P₂ - P₁
ΔP = 0.144 psi - 0.088 psi
<u>ΔP = 0.056 psi</u>
Answer:
Explanation:
This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is
which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):
and get
12.1 m
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:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;

where;
a is the angular acceleration and
r is the radius of the circular path

Determine time of the rotation;

Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration

The magnitude of total linear acceleration is given by;

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²