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
Explanation:
It is given that,
Mass of the car 1, 
Initial speed of car 1,
(east)
Mass of the car 2, 
Initial speed of car 2,
(north)
(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :




The magnitude of speed,

V = 12.22 m/s
(b) Let
is the direction the wreckage move just after the collision. It is given by :



Hence, this is the required solution.
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
(85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
Answer:
= 1,386 m / s
Explanation:
Rocket propulsion is a moment process that described by the expression
- v₀ =
ln (M₀ / Mf)
Where v are the velocities, final, initial and relative and M the masses
The data they give are the relative velocity (see = 2000 m / s) and the initial mass the mass of the loaded rocket (M₀ = 5Mf)
We consider that the rocket starts from rest (v₀ = 0)
At the time of burning half of the fuel the mass ratio is that the current mass is
M = 2.5 Mf
- 0 = 2000 ln (5Mf / 2.5 Mf) = 2000 ln 2
= 1,386 m / s
Answer:
No, both the thermometers will give the different reading.
Explanation:
Given,
- Both thermometer has same ice point =

- Both thermometer has same steam point =

- Distance between the ice point and steam point in both the thermometer is same of 100 division,
All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion. Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.