Answer:
0.0327 m
Explanation:
m = 2 kg
ω = 24 rad/s
A = 0.040 m
Let at position y, the potential energy is twice the kinetic energy.
The potential energy is given by
U = 1/2 m x ω² x y²
The kinetic energy is given by
K = 1/2 m x ω² x (A² - y²)
Equate both the energies as according to the question
1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)
y² = 2 A² - 2 y²
3y² = 2A²
y² = 2/3 A²
y = 0.82 A = 0.82 x 0.040 = 0.0327 m
Base on my research, within 2 hours you have a number of atoms which remain.
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0
So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0
It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms
N0= 5.55*10*8/0.2057 = 2.698*10^9 atoms
Therefore, 2.698*10^9 atoms is the number of N0
On Earth, none of the atmosphere is.
Answer:the maximum Hall voltage across the strip= 0.00168 V.
Explanation:
The Hall Voltage is calculated using
Vh= B x v x w
Where
B is the magnitude of the magnetic field, 5.6 T
v is the speed/ velocity of the strip, = 25 cm/s to m/s becomes 25/100=0.25m/s
and w is the width of the strip= 1.2 mm to meters becomes 1.2 mm /1000= 0.0012m
Solving
Vh= 5.6T x 0.25m/s x 0.0012m
=0.00168T.m²/s
=0.00168Wb/s
=0.00168V
Therefore, the maximum Hall voltage across the strip=0.00168V
Explanation:
the object has constant velocity for 2 seconds and it get a constant accelration (2ms-2)
