Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
Answer:
Explanation:
d = 42 meters
v = 12 m/s
t = ?
t = d/v
t = 42 / 12
t = 3.5 seconds. That's awfully fast.
Answer:
built in potential Vbi = +0.5V
barrier height = 0.139 V
doping concentration = 5.39 × 10²³cm³
Explanation:
Answer: a) 7.71 m/s², b) - 6.67 m/s²
Explanation: first thing to note is that
1 mile = 1609.34
1 hour = 3600s
Hence, 100mph to m/s = (100 × 1609.34)/3600 = 44.71 m/s
Initial velocity (u) = 0, final velocity (v) = 44.71 m/s, t = 5.8s, a = acceleration =?
By using newton's laws of motion
v = u + at
44.71 = 0 + a(5.8)
44.71 = 5.8a
a = 44.71/5.8
a = 7.71 m/s²
Question b)
The car is completing a stop which implies that the car is coming to rest, and when a car is coming to rest, the final velocity (v) is zero.
Hence u = 34 m/s, v = 0, a =?, t = 5.1 s
v = u + at
0 = 34 + a(5.1)
a(5.1) = - 34
a = - 34/5.1
a = - 6.67 m/s².
The negative sign beside the acceleration shows that the body is decelerating
1. 100,000% A. I simply googled C12 and it's the most stable Isotope of Carbon...
2. A as well.. 99% sure.. and so far my 99% sure has been 100% correct so ... :p
