False. Velocity is a vector and is measured in m/s (in SI, anyway).
<u>Answer:</u>
1) Distance traveled by bird = 403 meter
2)Average speed = 1.66 km /hour
3) Zcceleration = 2 
<u>Explanation:</u>
1) Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.
2) Average speed = Total distance covered / Time taken for that distance to cover.
Total distance covered = 2+0.5+2.5 = 5 km
Time taken = 3 hours
Average speed = 5/3 = 1.66 km /hour
3) Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.
Change in velocity = 14 - 6 = 8 m/s
Time = 4 seconds
So acceleration = 8 / 4 = 2 
Answer:
4.39 x 10^-19 J
Explanation:
q1 = 1.6 x 10^-19 C
q2 = - 1.6 x 10^-19 C
r1 = 3 x 10^-10 m
r2 = 7 x 10^-10 m
The formula for the potential energy is given by
U1 = k q1 q2 / r1 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (3 x 10^-10)
U1 = - 7.68 x 10^-19 J
U2 = k q1 q2 / r2 = - (9 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19) / (7 x 10^-10)
U2 = - 3.29 x 10^-19 J
Change in potential energy is
U2 - U1 = - 3.29 x 10^-19 + 7.68 x 10^-19 = 4.39 x 10^-19 J
Answer:
t = 0.55[sg]; v = 0.9[m/s]
Explanation:
In order to solve this problem we must establish the initial conditions with which we can work.
y = initial elevation = - 1.5 [m]
x = landing distance = 0.5 [m]
We set "y" with a negative value, as this height is below the table level.
in the following equation (vy)o is equal to zero because there is no velocity in the y component.
therefore:
![y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7Bo%7D%2At%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%5C%5C%20%20%20where%3A%5C%5C%28v_%7By%7D%29_%7Bo%7D%3D0%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%5Bsg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%5D%5C%5C%20-1.5%20%3D%200%2At%20-4.905%2At%5E%7B2%7D%20%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B1.5%7D%7B4.905%7D%20%7D%20%5C%5Ct%3D0.55%5Bs%5D)
Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.
![(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7B0.5%7D%7B0.55%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D0.9%5Bm%2Fs%5D)