Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
Considering that Work, W, is:
W=F·d
You have: 224=F·32
So that F=7 N
Answer:
hi everybody it's Satoshi you know that
Answer:
the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
Explanation:
Given;
energy of Mr. Voytko, E = 7.35 J
mass of the apple, m = 0.3 kg
Apply the principle of conservation of energy.
Energy of Mr. Voytko = Potential energy of the apple due to its height above the ground.
E = mgh
where;
h is the height above the ground through Mr. Voytko lifted the apple.
g is acceleration due to gravity = 9.8 m/s²
h = E / (mg)
h = 7.35 / (0.3 x 9.8)
h = 2.5 m
Therefore, the height above the ground through Mr. Voytko lifted the apple is 2.5 m.
Explanation:
It is given that,
Charge in the particle, ![q=6.4\ nC=6.4\times 10^{-9}\ C](https://tex.z-dn.net/?f=q%3D6.4%5C%20nC%3D6.4%5Ctimes%2010%5E%7B-9%7D%5C%20C)
Work done by the force, ![W_1=6.5\times 10^{-5}\ J](https://tex.z-dn.net/?f=W_1%3D6.5%5Ctimes%2010%5E%7B-5%7D%5C%20J)
The kinetic energy of the particle, ![E_k=3.7\times 10^{-15}\ J](https://tex.z-dn.net/?f=E_k%3D3.7%5Ctimes%2010%5E%7B-15%7D%5C%20J)
(a) Using the conservation of energy to find the work done by the electric force. Let it is given by
as :
![W_1+W_2=E_k](https://tex.z-dn.net/?f=W_1%2BW_2%3DE_k)
![W_2=E_k-W_1](https://tex.z-dn.net/?f=W_2%3DE_k-W_1)
![W_2=3.7\times 10^{-15}-6.5\times 10^{-5}](https://tex.z-dn.net/?f=W_2%3D3.7%5Ctimes%2010%5E%7B-15%7D-6.5%5Ctimes%2010%5E%7B-5%7D)
![W_2=-6.49\times 10^{-5}\ J](https://tex.z-dn.net/?f=W_2%3D-6.49%5Ctimes%2010%5E%7B-5%7D%5C%20J)
(b) Let V is the electric potential. the work done by the electric charge per unit positive charge is called its electric potential. Mathematically,
![\Delta V=\dfrac{W_2}{q}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cdfrac%7BW_2%7D%7Bq%7D)
![\Delta V=\dfrac{-6.49\times 10^{-5}}{6.4\times 10^{-9}}](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cdfrac%7B-6.49%5Ctimes%2010%5E%7B-5%7D%7D%7B6.4%5Ctimes%2010%5E%7B-9%7D%7D)
![\Delta V=10.14\times 10^3\ volts](https://tex.z-dn.net/?f=%5CDelta%20V%3D10.14%5Ctimes%2010%5E3%5C%20volts)
![\Delta V=10.14\ kV](https://tex.z-dn.net/?f=%5CDelta%20V%3D10.14%5C%20kV)
Hence, this is the required solution.