Answer: ![3.7 m/s^2](https://tex.z-dn.net/?f=3.7%20m%2Fs%5E2)
Explanation:
The acceleration of the motorcycle is given by Newton's second law:
![a=\frac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D)
where
F = 1295 N is the force applied to the motorcycle
m = 350 kg is the mass of the motorbike
a is the acceleration
By substituting the numbers into the formula, we find
![a=\frac{F}{m}=\frac{1295 N}{350 kg}=3.7 m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D%3D%5Cfrac%7B1295%20N%7D%7B350%20kg%7D%3D3.7%20m%2Fs%5E2)
Answer:
The net force acting on the otter along the incline is 13.96 N.
Explanation:
It is given that,
Mass of the otter, m = 2 kg
Distance covered by otter, d = 85 cm = 0.85 m
It takes 0.5 seconds.
We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :
![d=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=d%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Here, u = 0 (at rest)
![d=\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
![a=\dfrac{2d}{t^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B2d%7D%7Bt%5E2%7D)
![a=\dfrac{2\times 0.85}{0.5^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B2%5Ctimes%200.85%7D%7B0.5%5E2%7D)
![a=6.8\ m/s^2](https://tex.z-dn.net/?f=a%3D6.8%5C%20m%2Fs%5E2)
The net force acting on the otter along the incline is given by :
F = ma
![F=2\ kg\times 6.8\ m/s^2](https://tex.z-dn.net/?f=F%3D2%5C%20kg%5Ctimes%206.8%5C%20m%2Fs%5E2)
F = 13.6 N
So, the net force acting on the otter along the incline is 13.96 N. Hence, this is the required solution.
Answer:
Torque, ![\tau=3i+2j+4k](https://tex.z-dn.net/?f=%5Ctau%3D3i%2B2j%2B4k)
Explanation:
Given that,
Position of the flea, ![r=(-2i+j+k)\ m](https://tex.z-dn.net/?f=r%3D%28-2i%2Bj%2Bk%29%5C%20m)
Force acting on the flea, ![F=(-2j+k)\ N](https://tex.z-dn.net/?f=F%3D%28-2j%2Bk%29%5C%20N)
We need to find the net torque about the origin on a flea located at coordinates. Its formula is given by :
![\tau=r\times F](https://tex.z-dn.net/?f=%5Ctau%3Dr%5Ctimes%20F)
![\tau=(-2i+j+k) \times (-2j+k)](https://tex.z-dn.net/?f=%5Ctau%3D%28-2i%2Bj%2Bk%29%20%5Ctimes%20%28-2j%2Bk%29)
On solving the cross product of r and F, we get :
![\tau=3i+2j+4k](https://tex.z-dn.net/?f=%5Ctau%3D3i%2B2j%2B4k)
So, the net torque about the origin on a flea is
. Hence, this is the required solution.
Answer:
i = qω/2π
Explanation:
Since the charge moves in a circle, its velocity v = rω. where r is the radius of the circle and ω the angular frequency of the charge.
The distance it covers in one cycle is the circle's circumference,d = 2πr.
The time it takes to cover this distance, t = d/v = 2πr/rω = 2π/ω
We know that current i = charge/time = q/t = q/2π/ω = qω/2π
So, the average current, i = qω/2π