1.) Have your keys in hand before approaching or entering your car.
2. Be alert to other pedestrians and drivers.
3.) Search for signs of movement between, beneath and around objects to both sides of your vehicle
4.) check the spare tire for proper inflation
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
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.
A mixture that appears to contain only one substance is a. homogenous mixture.
Answer: B) 0.00337 m3.
Explanation:
Given data:
Mass of the ball = 10kg
Weight of the ball in air = 98N
Weight of the ball in water = 65N
Solution:
To get the Volume of the ball when submerged in water, we divide the weight of the ball in water with the difference in apparent weight by 9.8m/s^2.
= 98 - 65 / 9.8
= 33 / 9.8
= 3.37kg
The volume of the ball is 3.37kg
The density of water is 1kg per Liter.
So 3.37 kg of water would have a volume of 3.37 Liters.
Therefore the ball would have a volume of 3.37 Liters (or 0.00337 cubic meters).