Answer:
a) 4 289.8 J
b) 4 289.8 J
c) 6 620.1 N
d) 411 186.3 m/s^2
e) 6 620.1 N
Explanation:
Hi:
a)
The kinetic energy of the bullet is given by the following formula:
K = (1/2) m * v^2
With
m = 16.1 g = 1.61 x 10^-2 kg
v = 730 m/s
K = 4 289.8 J
b)
the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:
W = ΔK = Kf - Ki = 4 289.8 J
c)
The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).
If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:
W = F * L
Where F is the net force and L is the length of the barrel, that is:
F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N
d)
The acceleration can be found dividing the force by the mass:
a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2
e)
The force will have a magnitude equal to c) and direction along the barrel towards the exit

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

★ The speed of the hound and the hare

★ The speed of the hound and the hare = 25:18

As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
Now the distance travelled by the hare in it's 6 leaps..!
Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
B when is spins it means its rotating east and west directions
Answer:
Explanation:
General equation of the electromagnetic wave:
![E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%20E_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
where
Phase angle, 0
c = speed of the electromagnetic wave, 3 × 10⁸
wavelength of electromagnetic wave, 698 × 10⁻⁹m
E₀ = 3.5V/m
Electric field equation
![E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=E%28x%2C%20t%29%3D%203.5sin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CE%28x%2C%20t%29%3D%203.5sin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)
Magnetic field Equation
![B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%20B_0sin%5B%5Cfrac%7B2%5Cpi%7D%7B%5Clambda%7D%28x-ct%29%2B%5Cphi%20%5D)
Where B₀= E₀/c

![B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]](https://tex.z-dn.net/?f=B%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%5Cfrac%7B2%5Cpi%7D%7B6.98%5Ctimes10%5E%7B-7%7D%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6%7D%28x-3%5Ctimes%2010%5E8t%29%5D%5C%5C%5C%5CB%28x%2C%20t%29%3D%201.2%5Ctimes10%5E%7B-8%7Dsin%5B%7B9%20%5Ctimes%2010%5E6x-2.7%5Ctimes%2010%5E%7B15%7Dt%29%5D)