Answer:
Force is work done divided by distance
Explanation:
Work done to an object is given as rhe product of force and diatance moved, hence expressed as W=Fd where F is the applied force in Newtons, d is diatance moved in meters and W is work. The relationship between force and distance then is found by making either F or d the subject hence
hence we can say that distance is given by dividing work by force or alternatively Force is relates to distance by dividing work done by distance. Both are correct.
Incorrect statement in Dalton's original atomic theory are that atoms are indestructible and statement that all atoms of one element are same mass and characteristics.
Modern atomic theory is, of course, a little bit updated but original Daltons's theory is still remains. By modern atomic theory atoms can be destructed by nuclear reaction, but no with chemical reaction. Also there are different kinds of atoms in one element, their mass can be different.
Answer:
4.24nm
0.385eV
Explanation:
Maximum wavelength (λmax) :
λmax = ( hc) /Φ
h = plancks constant = 6.63 * 10^-34
c = speed of light = 3*10^8
1ev = 1.6 * 10^-19
Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19
λmax = [(6.63 * 10^-34) * (3 * 10^8)] / 4.688*10^-19
λmax = 19.89 * 10^-26 / 4.688*10^-19
λmax = 4.242 * 10^-7 m
λmax= 4.24nm
B.)
E = hc / eλ eV
λ = 3.75nm = 3.75 * 10^-7m = 375 *10^-9
E = (6.63 * 10^-34) * (3 * 10^8) / (1.6 * 10^-19) * (375 * 10^-9)
E = 19.89 * 10^-26 / 600 * 10^-28
E = 0.03315 * 10^-26 + 28
E = 0.03315 * 10^2
E = 3.315 eV
Stopping potential : (3.315 eV - 2.93eV) = 0.385eV
Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m + 
) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m + )v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 + )v / Δr
f = (v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f = (343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz
It’s to push an object in it’s direction of where it’s leading to.
