1 and 4 are tire.
2 and 3 are not.
As per the question, the mass of meteorite [ m]= 50 kg
The velocity of the meteorite [v] = 1000 m/s
When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.
We are asked to calculate the gain in kinetic energy of earth.
The kinetic energy of meteorite is calculated as -
![Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2](https://tex.z-dn.net/?f=Kinetic%5C%20energy%5C%20%5BK.E%5D%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
![=\frac{1}{2}50kg*[1000\ m/s]^2](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D50kg%2A%5B1000%5C%20m%2Fs%5D%5E2)

Here, J stands for Joule which is the S.I unit of energy.
Answer:
i) 24.5 m/s
ii) 30,656 m
iii) 89,344 m
Explanation:
Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer
i) Los parámetros dados son;
Altura inicial, s = 120 m
El tiempo en caída libre = 2.5 s
De la ecuación de caída libre, tenemos;
v = u + gt
Dónde:
u = Velocidad inicial = 0 m / s
g = Aceleración debida a la gravedad = 9.81 m / s²
t = Tiempo de caída libre = 2.5 s
Por lo tanto;
v = 0 + 9.8 × 2.5 = 24.5 m / s
ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación
s = u · t + 1/2 · g · t²
= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m
iii) La altura restante = 120 - 30.656 = 89.344 m.
Answer:
A
Explanation:
Opposite attract. Since electrons are negative they will be attracted to the protons because they a positive
I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:

Where V is the voltage between the plates and d is separation.
Voltage is by definition:

Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:

Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.