Not the answer but the answer will nit be B. hope yall find the answer
The car's rate of acceleration : a = 2.04 m/s²
<h3>Further explanation</h3>
Given
speed = 110 km/hr
time = 15 s
Required
The acceleration
Solution
110 km/hr⇒30.56 m/s
Acceleration is the change in velocity over time
a = Δv : Δt
Input the value :
a = 30.56 m/s : 15 s
a = 2.04 m/s²
With constant angular acceleration
, the disk achieves an angular velocity
at time
according to
![\omega=\alpha t](https://tex.z-dn.net/?f=%5Comega%3D%5Calpha%20t)
and angular displacement
according to
![\theta=\dfrac12\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac12%5Calpha%20t%5E2)
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
![21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}](https://tex.z-dn.net/?f=21.0%5C%2C%5Cmathrm%7Brad%7D%3D%5Cdfrac12%5Calpha%281.00%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%5Calpha%3D42.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D)
b. Under constant acceleration, the average angular velocity is equivalent to
![\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5Comega_f%2B%5Comega_i%7D2)
where
and
are the final and initial angular velocities, respectively. Then
![\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5Cleft%2842.0%5Cfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%7D2%3D42.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
c. After 1.00 s, the disk has instantaneous angular velocity
![\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega%3D%5Cleft%2842.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%3D42.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
d. During the next 1.00 s, the disk will start moving with the angular velocity
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to
![\theta=\omega_0t+\dfrac12\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_0t%2B%5Cdfrac12%5Calpha%20t%5E2)
which would be equal to
![\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cleft%2842.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%2B%5Cdfrac12%5Cleft%2842.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%5E2%3D63.0%5C%2C%5Cmathrm%7Brad%7D)
Answer:
The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.
Explanation:
Electrical force is the repulsive or attractive force of attraction between any two charged bodies. it is possible to calculate electric force on anything that runs on batteries or uses a plug. Its net effect can be explained using newtons law of motion like any force.
From the coulombs law; F = k (q1║q2)/ r²
Where k = 9 ×10∧9 N.m².C∧-2, r12= 0.285m, q2= +6μC and q1 = 4.86 nC
substituting values in the equation
the net electrical force acting on the +6ц C charge is 5.4 N.