3.375m/s is the final velocity of the car.
<h3>How do you find final velocity?</h3>
The final velocity depends on how large the acceleration is and the distance over which it acts.
Initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.
According to the question,
A toy car starts from the rest and accelerates
So the acceleration = 1.50m/s²
Time = 2.25s
![x=x_{0} + vt](https://tex.z-dn.net/?f=x%3Dx_%7B0%7D%20%20%2B%20vt)
![x = 0 + ( 1.50m/s^2*2.25s)](https://tex.z-dn.net/?f=x%20%3D%200%20%2B%20%28%201.50m%2Fs%5E2%2A2.25s%29)
![x = 3.375m/s](https://tex.z-dn.net/?f=x%20%3D%203.375m%2Fs)
The final velocity, of the car is 3.375 m/s.
Learn more about velocity here:brainly.com/question/18084516
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Answer:I have to say 56
Explanation: because it is going up by 8
Answer:
Light comes in different colors like radio, ultra violet, gamma-ray, etc, and they are invisible to the bare eye
Explanation:
Answer:
The maximum value of the induced magnetic field is
.
Explanation:
Given that,
Radius of plate = 30 mm
Separation = 5.0 mm
Frequency = 60 Hz
Suppose the maximum potential difference is 100 V and r= 130 mm.
We need to calculate the angular frequency
Using formula of angular frequency
![\omega=2\pi f](https://tex.z-dn.net/?f=%5Comega%3D2%5Cpi%20f)
Put the value into the formula
![\omega=2\times\pi\times60](https://tex.z-dn.net/?f=%5Comega%3D2%5Ctimes%5Cpi%5Ctimes60)
![\omega=376.9\ rad/s](https://tex.z-dn.net/?f=%5Comega%3D376.9%5C%20rad%2Fs)
When r>R, the magnetic field is inversely proportional to the r.
We need to calculate the maximum value of the induced magnetic field that occurs at r = R
Using formula of magnetic filed
![B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}](https://tex.z-dn.net/?f=B_%7Bmax%7D%3D%5Cdfrac%7B%5Cmu_%7B0%7D%5Cepsilon_%7B0%7DR%5E2%5CtimesV_%7Bmax%7D%5Ctimes%5Comega%7D%7B2rd%7D)
Where, R = radius of plate
d = plate separation
V = voltage
Put the value into the formula
![B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}](https://tex.z-dn.net/?f=B_%7Bmax%7D%3D%5Cdfrac%7B4%5Cpi%5Ctimes10%5E%7B-7%7D%5Ctimes8.85%5Ctimes10%5E%7B-12%7D%5Ctimes%2830%5Ctimes10%5E%7B-3%7D%29%5E2%5Ctimes100%5Ctimes376.9%7D%7B2%5Ctimes130%5Ctimes10%5E%7B-3%7D%5Ctimes5.0%5Ctimes10%5E%7B-3%7D%7D)
![B_{max}=2.901\times10^{-13}\ T](https://tex.z-dn.net/?f=B_%7Bmax%7D%3D2.901%5Ctimes10%5E%7B-13%7D%5C%20T)
Hence, The maximum value of the induced magnetic field is
.
Answer:
a) d = 6.0 m
Explanation:
Since car is accelerating at uniform rate then here we can say that the distance moved by the car with uniform acceleration is given as
![d = \frac{(v_f + v_i)}{2} \times t](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B%28v_f%20%2B%20v_i%29%7D%7B2%7D%20%5Ctimes%20t)
here we know that
![v_f = 10 m/s](https://tex.z-dn.net/?f=v_f%20%3D%2010%20m%2Fs)
![v_i = 0](https://tex.z-dn.net/?f=v_i%20%3D%200)
![t = 1.2 s](https://tex.z-dn.net/?f=t%20%3D%201.2%20s)
now we will have
![d = \frac{(10 + 0)}{2}\times 1.2](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B%2810%20%2B%200%29%7D%7B2%7D%5Ctimes%201.2)
![d = 5 \times 1.2](https://tex.z-dn.net/?f=d%20%3D%205%20%5Ctimes%201.2)
![d = 6.0 m](https://tex.z-dn.net/?f=d%20%3D%206.0%20m)