Frictional force between the object and the floor=5 N
Explanation:
power= 50 W
velocity= 10 m/s
power= force * velocity
50=F * 10
F=50/10
F=5 N
Thus the force of friction= 5 N
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Answer:
c
Explanation:
wavelength shorter means energy is higher
the wavelength
radio waves>microwave>infrared rays>gamma rays
Answer:
Distancia = 17,5 kilómetros.
Explanation:
Dados los siguientes datos;
Velocidad = 36 km/h
Tiempo = 0.5 horas
Para encontrar la distancia recorrida;
Distancia = velocidad * tiempo
Distancia = 35 * 0.5
Distancia = 17,5 kilómetros.
Por tanto, la distancia recorrida por el automóvil es de 17,5 kilómetros.
Answer:
D = 4 m
Explanation:
Speed of cart in air track v₁ = 0.5 m/s
Speed of cart moved when air is turned off v₂= 1 m/s
The distance travelled by the cart is d₁ = 1 m
Work done (W) = F x d
Work done is equal to the kinetic energy
F x d = 1/2mv²
velocity is directly proportional to distance
therefore,
v₁²/ v₂² = d₁ / d₂
d₂ = d₁v₂² / v₁²
= 1 m x (1 m /s)² / (0.5 m/s)²
= 4 m
