The threshhold for gamma rays is between 1 * 10^17 Hz and 9*10^17 Hz
Since this question goes much beyond that the answer is you much be working with a gamma ray. It is "Best" classified this way because nothing else will fit. The others are all much lower.
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ =
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L = v² / g
(1 -cos²)/ cos θ =
1 - cos² θ = cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x=
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ =
T cos θ = m g
resolved
tan θ =
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
Answer:
》How can a camera help someone see something he or she wouldn't normally be able to see?
- Although the human eye is able to observe fast events as they happen, it is not able to focus on a single point of time. We cannot freeze motion with our eyes. With a camera, however, so long as there is enough light, we can freeze motion. ... The camera can capture 'the moment', while your eye cannot.
》In what way is laser different from light produced by other sources?
-The major difference between laser light and light generated by white light sources (such as a light bulb) is that laser light is monochromatic, directional and coherent. ... Light from a conventional sources, such as a light bulb diverges, spreading in all directions.
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
Answer:
The time taken for the alpha particle is 5.301x10⁻⁷s
Explanation:
The centripetal force is equal:
The magnetic force is equal:
Matching both expressions:
Where
m = 6.64x10⁻²⁷kg
v = 4.8x10⁵m/s
B = 0.123 T
q = 1.6x10⁻¹⁹C
The time taken for the alpha particle is: