Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 ×
m
separation L = 0.5 mm = 0.5 ×
m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 ×
/ 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 ×
sin(0.234) = 1 λ
λ = 2042.02 ×
m
λ = 2042 nm
Answer:
Force constant, k = 653.3 N/m
Explanation:
It is given that,
Weight of the bag of oranges on a scale, W = 22.3 N
Let m is the mass of the bag of oranges,


m = 2.27 kg
Frequency of the oscillation of the scale, f = 2.7 Hz
We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :



k = 653.3 N/m
So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.
Answer:
A, B, and E
Explanation:
The springs are identical, and are compressed the same amount, so they have the same initial elastic potential energy. (E is true)
Energy is conserved, so the darts have the same amount of kinetic energy. (A is true, C is false)
The lighter dart has the same energy as the heavier dart. Since it has less mass, it must have a greater speed. (B is true, D is false)
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.