a = (dx / dt)²
Explanation: Unit of distance is m (metres) and unit of time is s (seconds) speed v is first derivative of distance x versus time:
v = dx / dt, unit is m/s. Acceleration is second derivative of
speed versus time a = (dx / dt)² = (dv/dt) , unit is m/s²
Answer:
![E = 9.66\times 10^{-6} N/C](https://tex.z-dn.net/?f=E%20%3D%209.66%5Ctimes%2010%5E%7B-6%7D%20N%2FC)
direction is Horizontal
Explanation:
As we know that the string is horizontal here
so the tension force in the string is due to electrostatic force on it
now we will have
![F = qE](https://tex.z-dn.net/?f=F%20%3D%20qE)
so here the force is tension force on it
![F = 6.57 \times 10^{-2} N](https://tex.z-dn.net/?f=F%20%3D%206.57%20%5Ctimes%2010%5E%7B-2%7D%20N)
![Q = 6.80 \times 10^3 C](https://tex.z-dn.net/?f=%20Q%20%3D%206.80%20%5Ctimes%2010%5E3%20C)
now we have
![6.57 \times 10^{-2} = (6.80 \times 10^3)E](https://tex.z-dn.net/?f=6.57%20%5Ctimes%2010%5E%7B-2%7D%20%3D%20%286.80%20%5Ctimes%2010%5E3%29E)
![E = 9.66\times 10^{-6} N/C](https://tex.z-dn.net/?f=E%20%3D%209.66%5Ctimes%2010%5E%7B-6%7D%20N%2FC)
direction is Horizontal
This question would be false
Answer: W = 294 J
Explanation: Solution:
Work is expressed as the product of force and the distance of the object.
W = Fd where F = mg
W= Fd
= mg d
= 15 kg ( 9.8 m/s²) ( 2m )
= 294 J
1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.
<u>Explanation</u>:
We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the
and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity:
![\text { Kinetic Energy }=\frac{1}{2} m v^{2}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Kinetic%20Energy%20%7D%3D%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E%7B2%7D)
Better understood from numerical example as given:
If a man A having weight 50 kg run with speed 5 m/s and another man B having 100 kg weight run with 2.5 m / s. Which man will have more K.E?
This can be solved as follows:
![\text { Kinetic Energy of } \mathrm{A}=\frac{1}{2} 50 \times 5^{2}=625 \mathrm{J}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Kinetic%20Energy%20of%20%7D%20%5Cmathrm%7BA%7D%3D%5Cfrac%7B1%7D%7B2%7D%2050%20%5Ctimes%205%5E%7B2%7D%3D625%20%5Cmathrm%7BJ%7D)
![\text { Kinetic Energy bf } \mathrm{B}=\frac{1}{2} 100 \times 2.5^{2}=312.5 \mathrm{J}](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Kinetic%20Energy%20bf%20%7D%20%5Cmathrm%7BB%7D%3D%5Cfrac%7B1%7D%7B2%7D%20100%20%5Ctimes%202.5%5E%7B2%7D%3D312.5%20%5Cmathrm%7BJ%7D)
It shows that man A will have more K.E.
Hence 1 kg ball can have more K.E than 100 kg ball by doubling velocity.