Which of the following is an example of a primary source?
1 answer:
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If the density of water does not vary and the vents range in depth from about 1500 m to 3200 m below the surface, then the gauge pressure at a 2452-m deep vent is 224.268 atm.
Calculation:
Step-1:
It is given that the vents range in depth from about 1500 m to 3200 m below the surface. If we are assuming that the density of water does not vary. Then it is required to calculate the gauge pressure at a 2452-m deep vent.
The gauge pressure at a particular depth of ocean water is calculated as:
Here is the density of water, P is the required pressure, h is the depth of water, and g is the gravitational acceleration.
Step-2:
Now we are substituting the values to calculate the pressure at the depth of 2452-m.
Learn more about gauge pressure here,
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Answer:
a) Maximum height = 36.6 m
b) Horizontal distance at which the ball lands = 166.1 m
c) x-component = 32 m/s. y-component = - 27 m/s
Explanation:
Please, see the attached figure for a description of the problem.
The velocity vector "v" of the cannonball has two components, a horizontal component, "vx", and a vertical component "vy". Notice that at the maximum height, the vertical component "vy" of the velocity vector is 0.
In the same way, the position vector "r" is composed by "rx", its horizontal component, and "ry", the vertical component.
The velocity vector "v" ad the position vector "r" at time "t" are given by the following equations:
v = (v0 * cos α, v0 * sin α + g * t)
r = (x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)
Where
v0 = magnitude of the initial velocity vector
α = launching angle
g = gravity acceleration (-9.8 m/s², because the y-axis points up)
t = time
x0 = initial horizontal position
y0 = initial vertical position
If we consider the origin of the system of reference as the point at which the cannonball leaves tha catapult, then, x0 and y0 = 0
a) We know that at maximum height, the vertical component of the vector "v" is 0, because the ball does not move up nor down at that moment (see figure). Then:
0 = v0 * sin α + g * t
-v0 * sin α / g = t
-40 m/s * sin 37° / -9.8 m/s² = t
t = 2.5 s
We can now calculate the position of the cannonball at time t=2.5 s to obtain the maximum height:
r = (x0 + v0 * t cos α, y0 + v0 * t * sin α + 1/2 * g * t²)
The max height is the magnitude of the vector ry max (see figure). The vector ry max is:
ry = (0, y0 + v0 t sin α + 1/2 g * t²)
magnitude of ry = )
Then:
max height = y0 + v0 * t * sin α + 1/2 * g * t²
max height = 0 m + 40 m/s * 2.5 s * sin 37° - 1/2* 9.8 m/s² * (2.5 s)² = 29.6 m
Since the ball leaves the catapult 7 m above the ground, the max height above the ground will be 29.6 m + 7 m = 36.6m
<u>max height = 36.6 m</u>
b) When the ball hits the ground, the position is given by the vector "r final" (see figure). The magnitude of "rx", the horizontal component of "r final", is the horizontal distance between the catapult and the wall.
r final = ( x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)
We know that the vertical component of "r final" is -7 (see figure).
Then, we can obtain the time when the the ball hits the ground:
y0 + v0 * t * sin α + 1/2 * g * t² = -7 m
0 m + 40 m/s * t * sin 37° + 1/2 g * t² = -7 m
7 m + 40 m/s * t * sin 37° + 1/2 (-9.8 m/s²) * t² = 0
7 m + 24.1 m/s * t - 4.9 m/s² * t² = 0
solving the quadratic equation:
t = 5.2 s (The negative solution is discarded).
With this time, we can calculate the value of the horizontal component of "r final"
Distance to the wall = |rx| = x0 + v0 t cos α
|rx| = 0m + 40 m/s * 5.2 s * cos 37° =<u> 166.1 m</u>
c) With the final time obtained in b) we can calculate the velocity of the ball:
v = (v0 * cos α, v0 * sin α + g * t)
v =(40 m/s * cos 37°, 40 m/s * sin 37° -9.8 m/s² * 5.2 s)
v =(32 m/s, -27 m)
x-component = 32 m/s
y-component = - 27 m/s
Answer:
b = 0.89
Explanation:
The given vector is,
A is a unit vector
We need to find the value of b.
For a unit vector, |A| = 1
So,
So, th value of b is 0.89.