The gravitational potential energy of the brick is 25.6 J
Explanation:
The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.
Near the surface of a planet, the gravitational potential energy is given by
where
m is the mass of the object
g is the strength of the gravitational field
h is the height of the object relative to the ground
For the brick in this problem, we have:
m = 8 kg is its mass
g = 1.6 N/kg is the strenght of the gravitational field on the moon
h = 2 m is the height above the ground
Substituting, we find:
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Answer:
a = 0.5 m/s²
Explanation:
Applying the definition of angular acceleration, as the rate of change of the angular acceleration, and as the seats begin from rest, we can get the value of the angular acceleration, as follows:
ωf = ω₀ + α*t
⇒ ωf = α*t ⇒ α = 
= 
The angular velocity, and the linear speed, are related by the following expression:
v = ω*r
Applying the definition of linear acceleration (tangential acceleration in this case) and angular acceleration, we can find a similar relationship between the tangential and angular acceleration, as follows:
a = α*r⇒ a = 0.067 rad/sec²*7.5 m = 0.5 m/s²
Answer:
57.94°
Explanation:
we know that the expression of flux
where Ф= flux
E= electric field
S= surface area
θ = angle between the direction of electric field and normal to the surface.
we have Given Ф= 78 
E=
S=
= 
=0.5306
θ=57.94°
Answer: 80m
Explanation:
Distance of balloon to the ground is 3150m
Let the distance of Menin's pocket to the ground be x
Let the distance between Menin's pocket to the balloon be y
Hence, x=3150-y------1
Using the equation of motion,
V^2= U^s + 2gs--------2
U= initial speed is 0m/s
g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s
40m/s is contant since U (the coin is at rest is 0) hence V =40m/s
Slotting our values into equation 2
40^2= 0^2 + 2 * 10* (3150-y)
1600 = 0 + 63000 - 20y
1600 - 63000 = - 20y
-61400 = - 20y minus cancel out minus on both sides of the equation
61400 = 20y
Hence y = 61400/20
3070m
Hence, recall equation 1
x = 3150 - 3070
80m
I hope this solve the problem.
La respuesta es "Un avion que vuela al norte con rapidez constante y altitud constante".
Para que la fuerza neta sea 0, la aceleracion debe ser 0, para esto la velocidad debe ser constante.
Para que la velocidad sea constante el objecto debe estar moviendo con rapidez (magnitud de la velocidad) constante y sin cambiar direccion; ya que la velocidad es un vector asi es que depende en magnitud y direccion.
En las demas opciones la magnitud de la velocidad (rapidez) cambia y/o la direccion.
