Answer:
5.5 × 10^14 Hz or s^-1
no orange light has less frequency so no photoelectric effect
Explanation:
hf = hf0 + K.E
HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s
f is frequency of incident photon and f0 is threshold frequency
hf0 = hf- k.E
6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20
6.63 × 10 ^-34 × f0 = 3.64158×10^-19
f0 = 3.64158×10^-19/ 6.63 × 10 ^-34
f0 = 5.4925 × 10^14
f0 =5.5 × 10^14 Hz or s^-1
frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light
Answer:
B will take 1.034 times the time of A from Boston to Hartford.
Explanation:
Let the distance from Boston to Hartford be S.
Person A drives at a constant speed of 55 mph for the entire trip,
Time taken by person A
Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.
Time taken by person B
Ratio of time of arrival of B to A
B will take 1.034 times the time of A from Boston to Hartford.
Answer:
4.9 cm
Explanation:
From Hook's Law,
F = ke......................... Equation 1
Where F= force, e = extension, k = spring constant.
Note: the Force acting on the the spring is the weight of the mass.
W = mg.
F = mg.................... Equation 2
Where m = mass, g = acceleration due to gravity
Substitute equation 2 into equation 1
mg = ke
make e the subject of the equation
e = mg/k............... Equation 3.
Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m
e = (2×9.8)/400
e = 19.6/400
e = 0.049 m
e = 4.9 cm
Answer:
0.3677181864 m
Explanation:
u = Velocity = 1.5 m/s
= Angle = 20°
y = -20 cm
Velocity components
Acceleration components
Time taken is 0.26088 seconds
The distance the beetle travels on the ground is 0.3677181864 m
A goes with 2 and B goes with 1.
