Complete question:
A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity after 10s ?
Answer:
the final velocity of the train is 4 m/s.
Explanation:
Given;
initial velocity of the train, u = 44 m/s
acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)
time of motion, t = 10 s
let the final velocity of the train = v
The final velocity of the train is calculated using the following kinematic equation;
v = u + at
v = 44 + (-4 x 10)
v = 44 - 40
v = 4 m/s
Therefore, the final velocity of the train is 4 m/s.
Saludos!
Respuesta:28,64 m/s.
Explicación:Datos:
Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²
El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.
Entonces tenemos que:

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.
Sustituyendo tenemos que:

Que tengas un buen día!
Answer:
R = 98304.75 m = 98.3 km
Explanation:
The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.
Density = Mass/Volume
Now, it is given that the density of Earth has become:
Density = 1 x 10⁹ kg/m³
Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg
Volume = 4/3πR³ (Volume of Sphere)
R = Radius of Earth = ?
Therefore,
1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]
4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)
R³ = (3/4)(5.97 x 10¹⁵ m³)/π
R = ∛[0.95 x 10¹⁵ m³]
<u>R = 98304.75 m = 98.3 km</u>
Answer:

Explanation:
A parallel-plate capacitors consist of two parallel plates charged with opposite charge.
Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.
The electric field between two infinite sheets with opposite charge is:

where
is the surface charge density, where
Q is the charge on the plate
A is the area of the plate
is the vacuum permittivity
In this problem:
- The side of one plate is
L = 19 cm = 0.19 m
So the area is

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

Place the object in an electronic balance and measure its mass.
Place a measured amount of water in the cylinder.
Place the object in the cylinder so that it’s fully submerged.
Measure the new level of the liquid and subtract the original level. This is equal to the volume of the object.
Density = mass / volume.