A group of elements are highly inactive and are used by chemists because of their chemical stability.
1 answer:
You might be interested in
Answer:
Explanation:
The mass of that science book...wow. In pounds that would be 35.2! Yikes!
Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is
Δx where
v is the final velocity, our unknown,
v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,
a is the pull of gravity which is always -9.8 m/s/s, and
Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:
so
and
v = 48 m/s
As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:
Δx and
144 = -19.6Δx so
Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:
120 - 7.3 = 112.7 m off the ground.
Answer:
a) You should position the cannon at 981 m from the wall.
b) You could position the cannon either at 975 m or 7.8 m (not recomended).
Explanation:
Please see the attached figure for a graphical description of the problem.
In a parabolic motion, the position of the flying object is given by the vector position:
r =( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
where:
r = position vector
x0 = initial horizontal position
v0 = module of the initial velocity vector
α = angle of lanching
y0 = initial vertical position
t = time
g = gravity acceleration (-9.8 m/s²)
The vector "r" can be expressed as a sum of vectors:
r = rx + ry
where
rx = ( x0 + v0 t cos α ; 0)
ry = (0 ; y0 + v0 t sin α + 1/2 g t²)
rx and ry are the x-component and the y-component of "r" respectively (see figure).
a) We have to find the module of r1 in the figure. Note that the y-component of r1 is null.
r1 = ( x0 + v0 t cos α ; y0 + v0 t sin α + 1/2 g t²)
Knowing the the y-component is 0, we can obtain the time of flight of the cannon ball.
0 = y0 + v0 t sin α + 1/2 g t²
If the origin of the reference system is located where the cannon is, the y0 and x0 = 0.
0 = v0 t sin α + 1/2 g t²
0 = t (v0 sin α + 1/2 g t) (we discard the solution t = 0)
0 = v0 sin α + 1/2 g t
t = -2v0 sin α / g
t = -2 * 100 m/s * sin 53° / (-9.8 m/s²) = 16.3 s
Now, we can obtain the x-component of r1 and its module will be the distance from the wall at which the cannon sould be placed:
x = x0 + v0 t cos α
x = 0 m + 100m/s * 16.3 s * cos 53
x = 981 m
The vector r1 can be written as:
r1 = (981 m ; 0)
The module of r1 will be:
<u>Then, the cannon should be placed 981 m from the wall.</u>
b) The procedure is the same as in point a) only that now the y-component of the vector r2 ( see figure) is not null:
r2y = (0 ; y0 + v0 t sin α + 1/2 g t² )
The module of this vector is 10 m, then, we can obtain the time and with that time we can calculate at which distance the cannon should be placed as in point a).
module of r2y = 10 m
10 m = v0 t sin α + 1/2 g t²
0 = 1/2 g t² + v0 t sin α - 10 m
Let´s replace with the data:
0 = 1/2 (-9.8 m/s² ) t² + 100 m/s * sin 53 * t - 10 m
0= -4.9 m/s² * t² + 79.9 m/s * t - 10 m
Solving the quadratic equation we obtain two values of "t"
t = 0.13 s and t = 16.2 s
Now, we can calculate the module of the vector r2x at each time:
r2x = ( x0 + v0 t cos α ; 0)
r2x = (0 m + 100m/s * 16.2 s * cos 53 ; 0)
r2x = (975 m; 0)
Module of r2x = 975 m
at t = 0.13 s
r2x = ( 0 m + 100m/s * 0.13 s * cos 53 ; 0)
r2x = (7.8 m ; 0)
module r2x = 7.8 m
You can place the cannon either at 975 m or at 7.8 m (see the red trajectory in the figure) although it could be dangerous to place it too close to the enemy fortress!
