How much does 1 by 2 kg of tomatoes cost please tell the answer

Given,

Mass of the vehicle = 100 kg.

Initial velocity (u) = 5 m/sec

Final Velocity (v) = 0 m/sec

Time taken to stop it (t) = 1/10 sec

To Find,

Given the initial conditions, the required fore in the opposite direction to stop the car.

Solution,

We can solve this numerical by the following process,

We know F=m*a

where F= The amount of applied force

            m= mass of the vehicle

             a= acceleration of the vehicle

(Here all of the physical quantities are in SI units, So we do not have to do any unit conversion.)

Here, [tex]a=\frac{v-u}{t}\\ a=\frac{0-5}{\frac{1}{10} }\\a=(-5)*10\\a=-50[/tex]

Hence, acceleration is -50 [tex]m/s^2[/tex]

Therefore, Required force, F=m*a

F = (100*(-50))

F = -5000 N

The negative sign signifies that the required force is applied in the opposite direction.

Hence, the required force is -5000 N.

Answer:

Hey mate

Explanation:

hope this helps you :)

Plantation agriculture is a type of commercial farming in which a single crop is grown for the entire year. This type of farming requires large amount of labour and capital. The crop production may be further processed on the farm itself where it is grown or in nearby factories or small scale industries.

Answer:

Threshold frequency is 1.2 × 10^15 Hz.

Explanation:

Threshold frequency can be calculated by using the formula,

threshold frequency = work function (in joule)/ planks constant.

So, here change 5ev into joule by multiplying with 1.6 ×10^-19J which will give 8×10^-19J(approx)

and use planks constant, h= 6.6 ×10^-34J (approx)

Put in the above formula and then.. you got your answer.

Answer:

The threshold frequency for the metal is 1.21×10¹⁵Hz.

Explanation:

The threshold frequency is defined as the minimum frequency of the incident radiation below which the photoelectric emission is not possible completely. It is given as,

[tex]\nu_o=\frac{E_o}{h}[/tex]          (1)

Where,

ν₀=thereshold frequency

E₀=work function in joules

h=planks constant=6.6×10⁻³⁴ m²kg s⁻¹

The values given in the question are,

[tex]E_o=5\times 1.6\times 10^-^1^9=8\times 10^-^1^9J[/tex]    (2)

By placing the value of E₀ and h in equation (1) we get;

[tex]\nu_o=\frac{8\times 10^-^1^9}{6.6\times 10^-^3^4}=1.21\times 10^1^5 Hz[/tex]

Hence, the threshold frequency for the metal is 1.21×10¹⁵Hz.