Answer:
The magnitude of angular acceleration of the motor is
.
Explanation:
Given that,
Initial angular velocity, ![\omega_i=90\ rad/s](https://tex.z-dn.net/?f=%5Comega_i%3D90%5C%20rad%2Fs)
Final angular velocity, ![\omega_f=30\ rad/s](https://tex.z-dn.net/?f=%5Comega_f%3D30%5C%20rad%2Fs)
Angular displacement, ![\theta=180\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D180%5C%20rad)
Let
is the magnitude of the angular acceleration of the motor. It can be calculated using third equation of rotational kinematics as :
![\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{30^2-90^2}{2\times 180}\\\\\alpha =-20\ rad/s^2](https://tex.z-dn.net/?f=%5Comega_f%5E2-%5Comega_i%5E2%3D2%5Calpha%20%5Ctheta%5C%5C%5C%5C%5Calpha%20%3D%5Cdfrac%7B%5Comega_f%5E2-%5Comega_i%5E2%7D%7B2%5Ctheta%7D%5C%5C%5C%5C%5Calpha%20%3D%5Cdfrac%7B30%5E2-90%5E2%7D%7B2%5Ctimes%20180%7D%5C%5C%5C%5C%5Calpha%20%3D-20%5C%20rad%2Fs%5E2)
So, the magnitude of angular acceleration of the motor is
.
Answer:
The answer is (d) I promise i'm not lying I just checked on my test
.
Explanation:
Answer:
Wavelength = 3.74 m
Explanation:
In order to find wavelength in "metres", we must first convert megahertz to hertz.
1 MHz = 1 × 10⁶ Hz
80.3 Mhz = <em>x</em>
<em>x </em>= 80.3 × 1 × 10⁶ = 8.03 × 10⁷ Hz
The formula between wave speed, frequency and wavelength is:
v = fλ [where v is wave speed, f is frequency and λ is wavelength]
Reorganise the equation and make λ the subject.
λ = v ÷ f
λ = (3 × 10⁸) ÷ (8.03 × 10⁷)
λ = 3.74 m [rounded to 3 significant figures]
A. CORRECT.
A very thin rotor decreases the effectiveness in the braking system. It cannot withstand heat and can cause quick wearing and fading of brake parts.
B. INCORRECT
Air in the hydraulic system can only make brake pedal soft and spongy. Pulsation in brake pedals is as a result of worn out or unevenly brake pads.
Answer:
![0.95i+0.95j](https://tex.z-dn.net/?f=0.95i%2B0.95j)
Explanation:
We are given that
Length of wire,L=14 cm=![\frac{14}{100}=0.14 m](https://tex.z-dn.net/?f=%5Cfrac%7B14%7D%7B100%7D%3D0.14%20m)
1 m=100 cm
Current,I=1.5 k A
Magnetic force,F=(-0.2i+0.2 J)N
Magnetic force,F=![L(I\times B)=0.14(1.5k\times(B_xi+B_yj+B_zk))](https://tex.z-dn.net/?f=L%28I%5Ctimes%20B%29%3D0.14%281.5k%5Ctimes%28B_xi%2BB_yj%2BB_zk%29%29)
![-0.2i+0.2j=0.14(1.5B_xj-B_yi)=0.21B_xj-0.21B_yi](https://tex.z-dn.net/?f=-0.2i%2B0.2j%3D0.14%281.5B_xj-B_yi%29%3D0.21B_xj-0.21B_yi)
By comparing on both sides
![0.2=0.21B_x](https://tex.z-dn.net/?f=0.2%3D0.21B_x)
![B_x=\frac{0.2}{0.21}0.95=0.95T](https://tex.z-dn.net/?f=B_x%3D%5Cfrac%7B0.2%7D%7B0.21%7D0.95%3D0.95T)
![0.21B_y=0.2](https://tex.z-dn.net/?f=0.21B_y%3D0.2)
![B_y=\frac{0.2}{0.21}=0.95 T](https://tex.z-dn.net/?f=B_y%3D%5Cfrac%7B0.2%7D%7B0.21%7D%3D0.95%20T)
When the current flows in positive x direction
Then,![F=0.2k N](https://tex.z-dn.net/?f=F%3D0.2k%20N)
![0.2k=0.14(1.5i\times (B_xi+B_yj+B_zk)](https://tex.z-dn.net/?f=0.2k%3D0.14%281.5i%5Ctimes%20%28B_xi%2BB_yj%2BB_zk%29)
![0.2k=0.21B_yk-0.21B_zj](https://tex.z-dn.net/?f=0.2k%3D0.21B_yk-0.21B_zj)
By comparing on both sides we get
![0.21B_z=0](https://tex.z-dn.net/?f=0.21B_z%3D0)
![B_z=0](https://tex.z-dn.net/?f=B_z%3D0)
Magnetic field,![B=B_xi+B_yj+B_zk=0.95i+0.95j](https://tex.z-dn.net/?f=B%3DB_xi%2BB_yj%2BB_zk%3D0.95i%2B0.95j)