Answer:
26 days
Explanation:
m = 9.4×1021 kg
r= 1.5×108 m
F = 1.1×10^ 19 N
We know Fc = 
==> 1.1 × 
= (9.4 × 
× 
) ÷ 1.5 × 
==> 1.1 × = × 6.26×
==> = 1.1 × ÷ 6.26×
==> = 0.17571885 × 
==> v= 0.419188323 × 
m/sec
==> v= 419.188322834 m/s
Putting value of r and v from above in ;
T= 2πr ÷ v
==> T= 2×3.14×1.5× ÷ 0.419188323 ×
==> T = 22.472× 100000 = 2247200 sec
but
86400 sec = 1 day
==> 2247200 sec= 2247200 ÷ 86400 = 26 days
Answer:
La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.
Explanation:
Asumamos que tanto el niño como el carrusel no tienen carga externa aplicada sobre aquellos, de modo que se puede aplicar el Principio de Conservación de la Cantidad de Movimiento Angular:
(1)
Donde:
- Masa del niño, medida en kilogramos.
- Velocidad lineal inicial del niño, medida en metros por segundo.
- Radio máximo del carrusel, medida en metros.
- Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.
- Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.
Si sabemos que 
, 
, 
y 
, tenemos que la velocidad angular final es:
La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.
The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0 → <em>n</em> = <em>mg</em> = 1876 N
where <em>n</em> is the magnitude of the normal force.
In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,
∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0 → <em>f</em> = 747 N
The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that
<em>f</em> = <em>µn</em> → <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40
Answer:
X= 700 Joules
Explanation:
The question asked about the efficiency of the work done.
The formula for efficiency is: Efficiency = (Useful output / input work) * 100%
The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.
Using the formula of efficiency,
20 = (140/X) * 100
So, we simply solve the above equation.
X= 140*100/20
X= 700 Joules
Period of a simple pendulum = 2π √(L/G)
(25 sec/15) = 2π √(L / 9.8 m/s²)
5/3 sec = 2π √(L/9.8 m/s²)
5 sec / 6π = √ (L/9.8 m/s²)
(5sec · √9.8m/s²) / 6π = √L
Square each side:
(25 s²) · (9.8 m/s²) / 36π² = L
L = (25 · 9.8) / (36 π²) meters
L = 0.69 meter
