Answer:
because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.
Answer:
Yes is large enough
Explanation:
We need to apply the second Newton's Law to find the solution.
We know that,
![F= ma](https://tex.z-dn.net/?f=F%3D%20ma)
And we know as well that
![a= \frac{v}{t}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7Bv%7D%7Bt%7D)
Replacing the aceleration value in the equation force we have,
![F= \frac{mv}{t}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7Bmv%7D%7Bt%7D)
Substituting our values we have,
![F= \frac{(0.060)(55)}{4*10^{-3}}](https://tex.z-dn.net/?f=F%3D%20%5Cfrac%7B%280.060%29%2855%29%7D%7B4%2A10%5E%7B-3%7D%7D)
![F=825N](https://tex.z-dn.net/?f=F%3D825N)
The weight of the person is then,
![W = mg](https://tex.z-dn.net/?f=W%20%3D%20mg)
![W = (60)(9.8) = 558N](https://tex.z-dn.net/?f=W%20%3D%20%2860%29%289.8%29%20%3D%20558N)
<em>We can conclude that force on the ball is large to lift the ball</em>
Answer:
a). Maximum Length L=0.929m
b). T=0.83 Hz or 1.2s
c). Longer, the effortless waling T=2.1 Hz or t=0.475s
d). t=1.2s V=0.774 ![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
t=0.475s V=1.95 ![\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
Length legs=L=1.1m
angle=50
the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image
Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length
a).
![sin(\alpha) =\frac{op}{h}](https://tex.z-dn.net/?f=sin%28%5Calpha%29%20%3D%5Cfrac%7Bop%7D%7Bh%7D)
![op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m](https://tex.z-dn.net/?f=op%3Dh%2Asin%28%5Cfrac%7B%5Calpha%20%7D%7B2%7D%29%5C%5C%20op%3D1.1m%2Asin%28%5Cfrac%7B50%7D%7B2%7D%29%5C%5Cop%3D0.464m)
Length of the step
L=0.464m*2
L=0.928m
b).
period=T
![T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B1%7D%7Btime%7D%3D%5Cfrac%7B1%7D%7Bt%7D%5C%5C%20T%3D%5Cfrac%7B1%7D%7B1.2s%7D%5C%5CT%3D0.83%20s%5E%7B-1%7D%5C%5C%20T%3D0.83Hz)
c).
![T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz](https://tex.z-dn.net/?f=T1%3D2%5Cpi%20%2A%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D%20%5C%5CT1%3D2%5Cpi%20%2A%5Csqrt%7B%5Cfrac%7B1.1m%7D%7B9.8%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%7D%7D%5C%5C%20T1%3D2.1%20Hz)
The period is the inverse of the time of the motion so, the T1 is faster that the T because
![t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1%7D%7BT1%7D%3Dt%3D%20%5Cfrac%7B1%7D%7B2.1%7D%3D0.47s)
d).
The speed is the relation between the distance with time so:
![Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}](https://tex.z-dn.net/?f=Vt%3D%5Cfrac%7B0.928m%7D%7B1.2s%7D%20%5C%5CVt%3D0.773%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5CVt%3D%5Cfrac%7B0.928m%7D%7B0.475s%7D%20%5C%5CVt%3D1.953%20%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
(a) p = 3.4 kg-m/s (b) 37.78 N.
Explanation:
Mass of a basketball, m = 0.4 kg
Initial velocity of the ball, u = -5.7 m/s (as it comes down so it is negative)
It rebounds upward at a speed of 2.8 m/s (as it rebounds so positive)
(a) Change in momentum = final momentum - initial momentum
p = m(v-u)
p = 0.4 (2.8-(-5.7))
p = 3.4 kg-m/s
(b) Impulse = change in momentum
Ft = 3.4
We have, t = 0.09 s
![F=\dfrac{3.4}{0.09}\\\\F=37.78\ N](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B3.4%7D%7B0.09%7D%5C%5C%5C%5CF%3D37.78%5C%20N)
Hence, this is the required solution.