Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
The answer is mass. I have to comment more than 20 characters.
Answer:
Option C. 210 J.
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Velocity (v) = 18 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Total Mechanical energy (ME) =?
Next, we shall determine the potential energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Height (h) = 12 m
Acceleration due to gravity (g) = 9.8 m/s²
Potential energy (PE) =?
PE = mgh
PE = 0.75 × 9.8 × 12
PE = 88.2 J
Next, we shall determine the kinetic energy of the plane. This can be obtained as follow:
Mass (m) = 0.75 Kg
Velocity (v) = 18 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.75 × 18²
KE = ½ × 0.75 × 324
KE = 121.5 J
Finally, we shall determine the total mechanical energy of the plane. This can be obtained as follow:
Potential energy (PE) = 88.2 J
Kinetic energy (KE) = 121.5 J
Total Mechanical energy (ME) =?
ME = PE + KE
ME = 88.2 + 121.5
ME = 209.7 J
ME ≈ 210 J
Therefore, the total mechanical energy of the plane is 210 J.
Answer:
Explanation:
I got everything but i. Don't know why but it's eluding me. So let's do everything but that.
a. PE = mgh so
PE = (2.5)(98)(14) and
PE = 340 J
b.
so
and
KE = 250 J
c. TE = KE + PE so
TE = 340 + 250 and
TE = 590 J
d. PE at 8.7 m:
PE = (2.5)(9.8)(8.7) and
PE = 210 J
e. The KE at the same height:
TE = KE + PE and
590 = KE + 210 so
KE = 380 J
f. The velocity at that height:
and
so
v = 17 m/s
g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:
590 = KE + PE and
PE = (2.5)(9.8)(11.6) so
PE = 280 then
590 = KE + 280 so
KE = 310 then
and
so
v = 16 m/s
h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:
and
26 = 0 + 9.8t and
26 = 9.8t so the time at 26 m/s is
t = 2.7 seconds. Now we use that in the equation for displacement:
Δx =
and filling in the time the object was at 26 m/s:
Δx = 0t +
so
Δx = 36 m
i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.