A uniform brick of length 21 m is placed over
1 answer:
Answer:
15.75 m
Explanation:
First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.
Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.
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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
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, 
, so we can write:
(1)
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find