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Rudiy27
2 years ago
12

A toy car that 0.12 m long is used to model the actions of an actual car that is 6m long. Which ratio shows the relationship bet

ween the sizes of the model and the actual car?
Physics
1 answer:
ra1l [238]2 years ago
3 0

Answer:

We know that a 0.12m long toy car is used as a model of a 6m long car.

We want to find the ratio that shows the relationship between both cars (the toy one and the real one)

The ratio that shows the relationship between both measures is just the quotient between both lengths, for example, if we take:

0.12m/6m = 0.02

This means that the length of the toy car is 0.02 times the length of the long car.

We also could have:

6m/0.12m = 50

This means that the length of the real car is 50 times the length of the toy car.

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An electron in a television tube is accelerated uniformly from rest to a speed of 8.4\times 10^7~\text{m/s}8.4×10 ​7 ​​ m/s over
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Answer:

P=3.42×10^-6 J/s

Explanation:

From the kinematics of motion with constant acceleration we know that :  

vf^2=vi^2+2*a(xf-xi)

Where :

• vf , vi, are the the final and the initial velocity of the electron  

• a is the acceleration of the electron  

• xf , xi are the final and the initial position of the electron .

Strategy for solving the problem : at first from the given information we calculate the acceleration of the electron.  

Givens: vf = 8.4 x 10^7 m/s , vi, = 0 m/s , xf = 0.025 m and xi = 0 m  

vf^2 =vi^2+2*a(xf-xi)

vf^2-vi^2=2*a(xf-xi)

2*a(xf-xi)= vf^2-vi^2

          a = (vf^2-vi^2)/2(xf-xi)

Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

  = 1.411 × 10^17

From the acceleration and the previous Eq. we can calculate the final velocity of the electron but a new position xf = 0.01 m  

so,

vf^2 =vi^2+2*a(xf-xi)

vf^2 =5.312× 10^7

From the following Eq. we can calculate the time elapsed in this motion .  

xf =xi+vi*t+1/2*a*t

xf =xi+vi*t+1/2*a*t

  t=√2(xf-xi)/a

 t=3.765×10^-10 s

now we can use the power P Eq.  

 P=W/Δt => ΔK/Δt  

Where: the work done W change the kinetic energy K of the electron ,

ΔK=Kf-Ki=>1/2*m*vf^2-1/2*m*vi^2

P=1/2*m*vf^2-1/2*m*vi^2/Δt

P=3.42×10^-6 J/s

6 0
3 years ago
PLz I Need Help PLZZZ ASAP
AURORKA [14]

#1

As we are increasing the frequency in the simulation the wavelength is decreasing

So if speed remains constant then wavelength and frequency depends inversely on each other

If we are in boat and and moving over very small wavelengths then these small wavelength will be encountered continuously by the boat in short interval of times

#2

As we are changing the amplitude in the simulation there is no change in the speed frequency and wavelength.

So amplitude is independent of all these parameter

Amplitude of wave will decide the energy of wave

So light of greater intensity is the light of larger amplitude

#3

In our daily life we deal with two waves

1 sound waves

2 light waves

3 0
2 years ago
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