Answer:21.18 m
Explanation:
Given
initial speed u=10 m/s
height of building h=22 m
time taken to complete 22 m
![h=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
initial vertical velocity =0
![22=\frac{1}{2}gt^2](https://tex.z-dn.net/?f=22%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![t=\sqrt{\frac{22\times 2}{g}}](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B22%5Ctimes%202%7D%7Bg%7D%7D)
![t=2.11 s](https://tex.z-dn.net/?f=t%3D2.11%20s)
Horizontal Distance moved
![R=u_x\times t](https://tex.z-dn.net/?f=R%3Du_x%5Ctimes%20t)
![R=10\times 2.11](https://tex.z-dn.net/?f=R%3D10%5Ctimes%202.11)
![R=21.18 m](https://tex.z-dn.net/?f=R%3D21.18%20m)
Answer:
The y-component of the electric force on this charge is ![F_y = -1.144\times 10^{-6}\ N.](https://tex.z-dn.net/?f=F_y%20%3D%20-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20N.)
Explanation:
<u>Given:</u>
- Electric field in the region,
![\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%28148.0%5C%20%5Chat%20i-110.0%5C%20%5Chat%20j%29%5C%20N%2FC.)
- Charge placed into the region,
![q = 10.4\ nC = 10.4\times 10^{-9}\ C.](https://tex.z-dn.net/?f=q%20%3D%2010.4%5C%20nC%20%3D%2010.4%5Ctimes%2010%5E%7B-9%7D%5C%20C.)
where,
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
![\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.](https://tex.z-dn.net/?f=%5Cvec%20E%20%3D%20%5Cdfrac%7B%5Cvec%20F%7D%7Bq%7D%5C%5C%5Ctherefore%20%5Cvec%20F%20%3D%20q%5Cvec%20E%5C%5C%3D%2810.4%5Ctimes%2010%5E%7B-9%7D%29%5Ctimes%20%28148.0%5C%20%5Chat%20i-110.0%5C%20%5Chat%20j%29%5C%5C%3D%281.539%5Ctimes%2010%5E%7B-6%7D%5C%20%5Chat%20i-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20%5Chat%20j%29%5C%20N.)
Thus, the y-component of the electric force on this charge is ![F_y = -1.144\times 10^{-6}\ N.](https://tex.z-dn.net/?f=F_y%20%3D%20-1.144%5Ctimes%2010%5E%7B-6%7D%5C%20N.)
Answer:
The speed of the riders on the Singapore Flyer is approximately 0.262 m/s
Explanation:
The dimensions of the tallest Ferris wheel in the world are;
The diameter of the Ferris wheel, D = 150 m
The tine it takes the Ferris wheel to make a full circle, T = 30 minutes = 30 min × 60 s/min = 1,800 seconds
The angular velocity of the Ferris wheel, ω = 2·π/T
The linear velocity of the Ferris wheel, v = r·ω = The speed of the riders
Where;
r = The radius of the Ferris wheel = D/2
D = 150 m
∴ r = 150 m/2 = 75 m
∴ v = r·2·π/T
∴ v = 75 m × 2 × π/(1,800 s) ≈ 0.262 m/s
The speed of the riders on the Singapore Flyer, v ≈ 0.262 m/s
Answer:
physical science
earth science and life science