A car traveling at 10 m/s passes over a hill on a road that has a circular cross section of radius 30 m. What is the force exert
1 answer:
You might be interested in
Answer:
he mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.
Explanation:
A radioactive material is transformed into another material by the emission of some particular radioactive ones, the most common being alpha and beta rays, which is why in the transformation process a certain amount of mass is lost. The process is described by the expression
N = No
From this expression the quantity half life time () is defined with time so that half of the atoms have been transformed
T_{1/2} = ln 2 /λ
in this case it does not indicate that T_{1/2}= 20 days is worth, for which periods have passed, in the first the number of radioactive atoms was reduced to half the number, leaving N´ and the second halved the number of nuclei that they were radioactive, leaving radioactive nuclei
first time of life
N´ = ½ N
second time of life
N´´ = ½ N´
N´´ = ¼ N
consequently in the sample at the end of these two decay periods we have, assuming that after each emission the atom is stable (non-radioactive). After the first emission there are n₁ = N / 2 stable atoms, after the second emission n₂ = ¼ N stable atoms are added and there are still n₃ = ¼ N radioactive atoms, so the total number of atoms is
n_total = n₁ + n₂ + n₃
Recall that the mass of the initial radioactive atoms is m₁, when transforming its mass of stable atoms is m₂ where
m₂ < m₁
therefore mass of
m_total = m₂ N / 2 + m₂ N / 4 + m₁ N / 4
m_total = m₂ ¾ N + m₁ ¼ N
m_total = N ( ¾ m₂ + ¼ m₁)
Since the mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.
Answer:
3.4 x 10⁴ m/s
Explanation:
Consider the circular motion of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of electron = 9.1 x 10⁻³¹ kg
v = radial speed
r = radius of circular path = 2 mm = 0.002 m
q = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
For the circular motion of electron
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Consider the linear motion of the electron :
v' = linear speed
x = horizontal distance traveled = 9 mm = 0.009 m
t = time taken = =
= 4.5 x 10⁻⁷ sec
using the equation
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
Speed is given as
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s