Answer:
Explanation:
Δ
- Δ
is the difference in velocity before and after a given time. 
is the acceleration of the object during this time. 
is time
is another way to write this equation.
- The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δ
- I rearranged this equation to solve for , but this is a step that you don't need to take, it's just good to get in the habit of doing this.
- Plug in the given values. Note that our final velocity is
, because the car travels until at <em>rest</em>.
![a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv_f-v_i%7D%7Bt%7D%5C%5Ca%3D%5Cfrac%7B%280%29-%5B%2817.1%5Cfrac%7Bmiles%7D%7Bhour%7D%20%29%28%5Cfrac%7Bhour%7D%7B3600s%7D%29%28%5Cfrac%7B1609.34m%7D%7Bmile%7D%29%5D%7D%7B9.7s%7D)
- Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.
![a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%280%29-%5B%2817.1%5Cfrac%7Bmiles%7D%7Bhour%7D%20%29%28%5Cfrac%7Bhour%7D%7B3600s%7D%29%28%5Cfrac%7B1609.34m%7D%7Bmile%7D%29%5D%7D%7B9.7s%7D%5C%5Ca%3D%5Cfrac%7B0m%2Fs-7.6m%2Fs%7D%7B9.7s%7D%20%5C%5Ca%3D%5Cfrac%7B-7.6m%2Fs%7D%7B9.7s%7D)
- if you converted correctly, your answer for
will be ≅ 
. - Now divide. Notice that the units for acceleration are
or <em>meters per second, per second</em>.
- Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.
<span>Landforms include striations, cirques, glacial horns, arêtes, trim lines, U-shaped valleys, </span>roches moutonnées<span>, overdeepenings and hanging valleys. These are all created by Glaciers!
</span>
Answer:
Line 1: the spacing is even the whole time
Line 2: the spacing increases over time
Line 3: the spacing decreases over time
hope this helps!
Answer:
313.6 m downward
Explanation:
The distance covered by the bullet along the vertical direction can be calculated by using the equation of motion of a projectile along the y-axis.
In fact, we have:
where
y(t) is the vertical position of the projectile at time t
h is the initial height of the projectile
is the initial vertical velocity of the projectile, which is zero since the bullet is fired horizontally
t is the time
a = g = -9.8 m/s^2 is the acceleration due to gravity
We can rewrite the equation as
where the term on the left, 
, represents the vertical displacement of the bullet. Substituting numbers and t = 8 s, we find
So the bullet has travelled 313.6 m downward.
Responder:
Velocidad = 41.5m / s
Espacio recorrida = 352.5 metros
Explicación:
Dado lo siguiente:
Velocidad inicial (u) = 19.8 km / h
Aceleración (a) = 2.4m / s ^ 2
Tiempo de viaje (t) = 15 s
A.) velocidad después de 15 s
Velocidad inicial = (19.8 × 1000) m / 3600s Velocidad inicial = 19800m / 3600 = 5.5m / s
Usando la ecuación: v = u + at, donde v es la velocidad
v = 5.5 + 2.4 (15)
v = 5.5 + 36
v = 41.5m / s
Espacio recorrida:
v ^ 2 = u ^ 2 + 2aS; donde S es la distancia recorrida
41.5 ^ 2 = 5.5 ^ 2 + 2 × (2.4) × S
1722.25 = 30.25 + 4.8S
1722.25 - 30.25 = 4.8S
1692 = 4.8S S = 1692 / 4.8 S = 352.5m
