Explain how electromagnetic waves are produced. I

A car mass 600kg starts from rest moving uniform acceleration 0.2 m/s^2 after 60 seconds collides with stationary pick up van of

nlexa [21]

Answer:

The given phenomenon supports the Principle of Conservation of momentum.

Explanation:

law of conservation of momentum

Initial Momentum = Final Momentum

So, first we calculate initial momentum of the system:

Initial Momentum = m1*u1 + m2*u2

and we are given

m1 = mass of car = 600 kg

m2 = mass of van = 400 kg

u1 = Initial Speed of Car

to find the initial speed we use equation of motion

Vf = Vi + at

Vf = 0 m/s + (0.2 m/s²)(60 s)

Vf = u₁ = 12 m/s

u₂ = Initial Speed of Van = 0 m/s

Therefore,

Initial Momentum = (600 kg)(12 m/s) + (400 kg)(0 m/s)

Initial Momentum = 7200 Ns .........(1)

final momentum = m₁v₁ + m₂v₂

where,

v₁ = v₂ = final speed of car+van (both locked ) = 7.2 m/s

Therefore,

Final Momentum = (600 kg)(7.2 m/s) + (400 kg)(7.2 m/s)

Final Momentum = 7200 Ns -------- (2)

on comparing (1) and (2)

Initial momentum = Final Momentum

Hence, the phenomenon of the system supports the principle of conservation of momentum.

6 0

3 years ago

A 30kg uniform solid cylinder has a radius of 0.18m. if the cylinder accelerates at 0.023 rad/s^2 as it rotates about an axis th

Nuetrik [128]

Answer:

0.011 N-m

Explanation:

Given that

The mass of a solid cylinder, m = 30 kg

The radius of the cylinder, r = 0.18 m

The acceleration of the cylinder, $\alpha =0.023\ rad/s^2$

It rotates about an axis through its center. We need to find the torque acting on the cylinder. The formula for the torque is given by :

$\tau=I\alpha$

Where

I is the moment of inertia of the cylinder,

For cylinder,

$I=\dfrac{mr^2}{2}$

So,

$\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{30\times (0.18)^2\times 0.023 }{2}\\\\\tau=0.011\ N-m$

So, the required torque on the cylinder is 0.011 N-m.

7 0

3 years ago

A juggler throws balls into the air. He throws one when the previous one is at its highest point. How high the balls rise if he

VikaD [51]

<span>If "m" balls are thrown per second, the time taken for a ball to reach its maximum height will be 1/m seconds. How to get this? See that the next ball is thrown only when the previous ball reaches its maximum height. If 'm' balls were thrown in 1 second this means that each ball was attaining its maximum ht in 1/m seconds. This was the main part. Now we can proceed to find maximum height in 2 ways- a) We know for upward journey ,
t=1/m
a=-g
v=u-gt
final velocity ,v = 0 (at highest point) u
=gt = g/m
Now we can apply h=ut-1/2 gt^2
Putting the values of u,t, we will get h= g/2m^2
b) The second method uses a trick that time taken to reach the maximum ht is same as time taken to fall down. So, we will now consider the downward journey of ball which also takes 1/m seconds We apply
h=ut+1/2gt^2
Here u=0 ,t=1/m
We will again get ,
h=g/2m^2</span>

6 0

3 years ago

Explain why air masses do not mix

kati45 [8]

Different densities have to have a reason - different pressure and/or humidity etc. If there is a different pressure, there is a mechanical force that preserves the pressure difference: think about the cyclones that have a lower pressure in the center. The cyclones rotate in the right direction and the cyclone may be preserved by the Coriolis force.

If the two air masses differ by humidity, the mixing will almost always lead to precipitation - which includes a phase transition for water etc. It's because the vapor from the more humid air mass gets condensed under the conditions of the other. You get some rain. In general, intense precipitation, thunderstorms, and other visible isolated weather events are linked to weather fronts.

At any rate, a mixing of two air masses is a nontrivial, violent process in general. That's why the boundary is called a "front". In the military jargon, a front is the contested frontier of a conflict. So your idea that the air masses could mix quickly and peacefully - whatever you exactly mean quantitatively - either neglects the inertia of the air, a relatively low diffusion coefficient, a low thermal conductivity, and/or high latent heat of water vapor. A front is something that didn't disappear within minutes so pretty much tautologically, there must be forces that make such a quick disappearance impossible.

3 0

3 years ago

In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$