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:
C. 10kg to 10kg
Explanation:
You have to picture to it I think
Answer:
(B) False
Explanation:
No, it is not possible to have thunder without lightning. Thunder is a direct result of lightning.
Answer:
Explanation:
distance travelled
s = 2πR
= 2 X 3.14 X 140
= 880 m
final velocity = v
initial velocity = u
distance travelled = s
time = 60 s
s = ut + 1/2 at²
880 = .5 x a x 60²
a = .244 m/s²
final velocity v = at
= .244 x 60
= 14.66
centripetal acceleration at final moment
v² /R
14.66 X 14.66 / 140
= 1.53 m/s⁻²
1.53 m/s²
this is centripetal acceleration which acts towards the centre.
tangential acceleration calculated a _t = .244
redial acceleration ( centripetal ) = 1.53
Resultant acceleration
R²= 1.53² + .244 ²
R = 1.55 m/s²
total force = 1.55 x 76
= 118 N
Hopefully I’m not late and I apologize if I am, but the answer to your question would be 95.6 km/hr. You know you can look up your question as well to see if they already have a answer to that so you won’t waste your points.
