Answer:
The football leaves with the velocity, u = 15.68 m/s
Explanation:
Given data,
The football bounces back up off the ground and is airborne for, t = 3.2 s
Let the football bounces back up off the ground in the vertical direction
The formula for time of flight is given by,
t = 2u /g
∴ u = gt / 2
Substituting the values,
u = 9.8 x 3.2 / 2
u = 15.68 m/s
Hence, the football leaves with the velocity, u = 15.68 m/s
Answer:
The minimum speed when she leave the ground is 6.10 m/s.
Explanation:
Given that,
Horizontal velocity = 1.4 m/s
Height = 1.8 m
We need to calculate the minimum speed must she leave the ground
Using conservation of energy
![K.E+P.E=P.E+K.E](https://tex.z-dn.net/?f=K.E%2BP.E%3DP.E%2BK.E)
![\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv_%7B1%7D%5E2%2B0%3Dmgh%2B%5Cdfrac%7B1%7D%7B2%7Dmv_%7B2%7D%5E2)
![\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7B2%7D%3Dgh%2B%5Cdfrac%7Bv_%7B2%7D%5E2%7D%7B2%7D)
Put the value into the formula
![\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7B2%7D%3D9.8%5Ctimes1.8%2B%5Cdfrac%7B%281.4%29%5E2%7D%7B2%7D)
![\dfrac{v_{1}^2}{2}=18.62](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_%7B1%7D%5E2%7D%7B2%7D%3D18.62)
![v_{1}=\sqrt{2\times18.62}](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Csqrt%7B2%5Ctimes18.62%7D)
![v_{1}=6.10\ m/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D6.10%5C%20m%2Fs)
Hence, The minimum speed when she leave the ground is 6.10 m/s.
Answer:
The component of waves must have same frequency and phase.
Explanation:
when the component of waves vibrate at the same rate and attain maximum point at the same time, reinforcement of the waves amplitude occur to cause a constructive interference.However, when the two waves are out of phase where one is at minimum when the other is at maximum a destructive interference happens.
The force between two charged particles is an electrostatic force. The intensity of this force is given by the following formula:
![F_c=k_c\frac{q_1q_2}{r^2}](https://tex.z-dn.net/?f=F_c%3Dk_c%5Cfrac%7Bq_1q_2%7D%7Br%5E2%7D)
Where q1 and q2 are charges r is the distance, and k_c is Coulomb's constant.
Coulomb's constant has the value of:
![k_c=8.9875\cdot 10^9\frac{Nm^2}{C^2}](https://tex.z-dn.net/?f=k_c%3D8.9875%5Ccdot%2010%5E9%5Cfrac%7BNm%5E2%7D%7BC%5E2%7D)
When we plug all the given values into the formula we get:
![F_c=8.9875\cdot 10^9\cdot\frac{-2\cdot3}{80^2}=-8425781 $N](https://tex.z-dn.net/?f=F_c%3D8.9875%5Ccdot%2010%5E9%5Ccdot%5Cfrac%7B-2%5Ccdot3%7D%7B80%5E2%7D%3D-8425781%20%24N)
Negative sign means that force is attractive.