The distance an object falls, from rest, in gravity is
D = (1/2) (G) (T²)
'T' is the number seconds it falls.
In this problem,
0.92 meter = (1/2) (9.8) (T²)
Divide each side by 4.9 : 0.92 / 4.9 = T²
Take the square root
of each side: √(0.92/4.9) = T
0.433 sec = T
The horizontal speed doesn't make a bit of difference in
how long it takes to reach the floor. BUT ... if you want to
know how far from the table the pencil lands, you can find
that with the horizontal speed.
The pencil is in the air for 0.433 second.
In that time, it travels
(0.433s) x (1.4 m/s) = 0.606 meter
from the edge of the table.
Answer:
The initial speed of the cork was 1.57 m/s.
Explanation:
Hi there!
The equation of the horizontal position of the cork in function of time is the following:
x = x0 + v0 · t · cos θ
Where:
x = horizontal position at time t.
x0 = initial horizontal position.
v0 = initial speed of the cork.
t = time.
θ = launching angle.
If we place the origin of the frame of reference at the launching point, then x0 = 0.
We know that at t = 1.25 s, x = 1.50 m. We also know the launching angle so we can solve the equation of horizontal position for the initial speed, v0:
x = v0 · t · cos θ
x / t · cos θ = v0
v0 = 1.50 m / (1.25 s · cos (40.0°)
v0 = 1.57 m/s
The initial speed of the cork was 1.57 m/s.
Energy in a spring:
E = 0.5 * k * x²
k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m
E = 0.5 * 800 * 0.05² = 1
Answer with Explanation:
We are given that
Diameter of fighter plane=2.3 m
Radius=![r=\frac{d}{2}=\frac{2.3}{2}=1.15 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B2.3%7D%7B2%7D%3D1.15%20m)
a.We have to find the angular velocity in radians per second if it spins=1200 rev/min
Frequency=![\frac{1200}{60}=20 Hz](https://tex.z-dn.net/?f=%5Cfrac%7B1200%7D%7B60%7D%3D20%20Hz)
1 minute=60 seconds
Angular velocity=![\omega=2\pi f](https://tex.z-dn.net/?f=%5Comega%3D2%5Cpi%20f)
Angular velocity=![2\times \frac{22}{7}\times 20=125.7 rad/s](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B22%7D%7B7%7D%5Ctimes%2020%3D125.7%20rad%2Fs)
b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.
![v=r\omega=1.15\times 125.7=144.56 m/s](https://tex.z-dn.net/?f=v%3Dr%5Comega%3D1.15%5Ctimes%20125.7%3D144.56%20m%2Fs)
c.Centripetal acceleration=![\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2](https://tex.z-dn.net/?f=%5Comega%5E2%20r%3D%28125.7%29%5E2%281.15%29%3D18170.56%20m%2Fs%5E2)
Centripetal acceleration==![\frac{18170.56\times g}{9.81}=1852.25 g m/s^2](https://tex.z-dn.net/?f=%5Cfrac%7B18170.56%5Ctimes%20g%7D%7B9.81%7D%3D1852.25%20g%20m%2Fs%5E2)