The rings of Saturn occupy the region inside Saturn's Roche limit.
1 answer:
You might be interested in
#1
A point on the outside of a spinning top whose rotational speed is constant.
Since the point at the top is at rest so it will not accelerate at all
A skydiver whose air resistance is equal to that of her weight.
Since weight is counter balanced by air resistance so net force on it is zero so acceleration will be zero
A car on the freeway experiencing a net force of -120 N.
Since net force is not zero so it will accelerate here
A submerged beach ball whose buoyant force is eight times the force of gravity on it.
Since buoyant force is more than the weight so it will not at rest and move upwards with some acceleration
#2
A frictionless spinning merry go round.
Since spinning wheel accelerating due to centripetal force so it is non inertial frame
A falling rock.
Falling rock is accelerating due to gravity so it is again non inertial frame
A hot air balloon moving at 30 degrees east of north with no net force.
since hot air balloon is having no force on it so its acceleration is zero and its an inertial frame
A space shuttle whose boosters just ignited for takeoff.
Since space shuttle is accelerating due to boosters so it is non inertial frame
#3
Here net force on it is due to two given forces
towards right
towards left
so here net force is given as
since force towards left is more so the direction will be towards left
so correct answer is 60 N towards left
Let the rise in temperature be
The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.
Here l = 93 m, α =
, and Δt =
So expansion in length = 93**5 = 0.007905 m =
So order of magnitude in change in length = -3
Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s