The galaxies aren't distributed randomly throughout the universe but they are grouped in gravitationally bound clusters.
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Answer:
A. something pushes or pulls it to stop.
Explanation:
Newtons first law. hope this helps
Answer:
a) ![v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Cfrac%7B%2862.5-66%29ft%7D%7B%282.5-2%29s%7D%3D-7ft%2Fs)
![v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D%5Cfrac%7B%2865.94-66%29ft%7D%7B%282.1-2%29s%7D%3D-0.6ft%2Fs)
![v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s](https://tex.z-dn.net/?f=v_%7B3%7D%3D%5Cfrac%7B%2866.0084-66%29ft%7D%7B%282.01-2%29s%7D%3D0.84ft%2Fs)
![v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s](https://tex.z-dn.net/?f=v_%7B4%7D%3D%5Cfrac%7B%2866.001-66%29ft%7D%7B%282.001-2%29s%7D%3D1ft%2Fs)
b) ![v=65-32(2)=1ft/s](https://tex.z-dn.net/?f=v%3D65-32%282%29%3D1ft%2Fs)
Explanation:
From the exercise we got the ball's equation of position:
![y=65t-16t^{2}](https://tex.z-dn.net/?f=y%3D65t-16t%5E%7B2%7D)
a) To find the average velocity at the given time we need to use the following formula:
![v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7By_%7B2%7D-y_%7B1%7D%20%20%7D%7Bt_%7B2%7D-t_%7B1%7D%20%20%7D)
Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001
![y_{t=2}=65(2)-16(2)^{2} =66ft](https://tex.z-dn.net/?f=y_%7Bt%3D2%7D%3D65%282%29-16%282%29%5E%7B2%7D%20%3D66ft)
![y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.5%7D%3D65%282.5%29-16%282.5%29%5E%7B2%7D%20%3D62.5ft)
![v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Cfrac%7B%2862.5-66%29ft%7D%7B%282.5-2%29s%7D%3D-7ft%2Fs)
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![y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.1%7D%3D65%282.1%29-16%282.1%29%5E%7B2%7D%20%3D65.94ft)
![v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s](https://tex.z-dn.net/?f=v_%7B2%7D%3D%5Cfrac%7B%2865.94-66%29ft%7D%7B%282.1-2%29s%7D%3D-0.6ft%2Fs)
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![y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.01%7D%3D65%282.01%29-16%282.01%29%5E%7B2%7D%20%3D66.0084ft)
![v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s](https://tex.z-dn.net/?f=v_%7B3%7D%3D%5Cfrac%7B%2866.0084-66%29ft%7D%7B%282.01-2%29s%7D%3D0.84ft%2Fs)
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![y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft](https://tex.z-dn.net/?f=y_%7Bt%3D2.001%7D%3D65%282.001%29-16%282.001%29%5E%7B2%7D%20%3D66.001ft)
![v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s](https://tex.z-dn.net/?f=v_%7B4%7D%3D%5Cfrac%7B%2866.001-66%29ft%7D%7B%282.001-2%29s%7D%3D1ft%2Fs)
b) To find the instantaneous velocity we need to derivate the equation
![v=\frac{df}{dt}=65-32t](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bdf%7D%7Bdt%7D%3D65-32t)
![v=65-32(2)=1ft/s](https://tex.z-dn.net/?f=v%3D65-32%282%29%3D1ft%2Fs)
Answer:
55000000 years
Explanation:
Rate of North American Plate moving (Velocity) = 20 mm/yr
Degrees New York has to move = 10° west
New York's longitude = 110 km/°
Distance of New York = 110×10 = 1100 km
= 1100×10⁶ mm
![\text {Time taken by the continental plate}=\frac {\text {Distance of New York}}{\text {Rate of North American Plate moving (Velocity)}}\\\Rightarrow \text {Time taken by the continental plate}=\frac{1100\times 10^6}{20}\\\Rightarrow \text {Time taken by the continental plate}=55\times 10^6\ years](https://tex.z-dn.net/?f=%5Ctext%20%7BTime%20taken%20by%20the%20continental%20plate%7D%3D%5Cfrac%20%7B%5Ctext%20%7BDistance%20of%20New%20York%7D%7D%7B%5Ctext%20%7BRate%20of%20North%20American%20Plate%20moving%20%28Velocity%29%7D%7D%5C%5C%5CRightarrow%20%5Ctext%20%7BTime%20taken%20by%20the%20continental%20plate%7D%3D%5Cfrac%7B1100%5Ctimes%2010%5E6%7D%7B20%7D%5C%5C%5CRightarrow%20%5Ctext%20%7BTime%20taken%20by%20the%20continental%20plate%7D%3D55%5Ctimes%2010%5E6%5C%20years)
∴ Time taken by New York to move 10° west is 55000000 years