Answer:
2.52 m/s
Explanation:
When the man takes a step, his foot is stationary while his body revolves around it. At the point when his body is directly above his foot, there will be no normal force at his maximum speed.
Sum of the forces in the radial direction:
∑F = ma
mg = m v² / r
g = v² / r
v = √(gr)
Given that r = 0.650 m:
v = √(9.8 m/s² × 0.650 m)
v = 2.52 m/s
Answer:
C
Sign-Negative
Explanation:
We are given that
Electric field =
(Radially downward)
Acceleration=
(Upward)
Mass of charge=3 g=
kg
1kg=1000g
We have to find the magnitude and sign of charge would have to be placed on a penny .
By newton's second law
![\sum F_y=ma](https://tex.z-dn.net/?f=%5Csum%20F_y%3Dma)
![\sum F_y=qE-mg](https://tex.z-dn.net/?f=%5Csum%20F_y%3DqE-mg)
Substitute the values then we get
![qE-mg=ma](https://tex.z-dn.net/?f=qE-mg%3Dma)
Substitute the values then we get
![q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)](https://tex.z-dn.net/?f=q%28100%29-3%5Ctimes%2010%5E%7B-3%7D%289.8%29%3D3%5Ctimes%2010%5E%7B-3%7D%280.19%29)
![100q-29.4\times 10^{-3}=0.57\times 10^{-3}](https://tex.z-dn.net/?f=100q-29.4%5Ctimes%2010%5E%7B-3%7D%3D0.57%5Ctimes%2010%5E%7B-3%7D)
![100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}](https://tex.z-dn.net/?f=100q%3D0.57%5Ctimes%2010%5E%7B-3%7D%2B29.4%5Ctimes%2010%5E%7B-3%7D%3D29.97%5Ctimes%2010%5E%7B-3%7D)
![q=\frac{29.97\times 10^{-3}}{100}](https://tex.z-dn.net/?f=q%3D%5Cfrac%7B29.97%5Ctimes%2010%5E%7B-3%7D%7D%7B100%7D)
C
Sign of charge =Negative
Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.
B) a rock being tossed high into the air
<span><em>Density</em>-dependent <em>factors</em> operate only when the population <em>density</em> reaches a certain level. </span>