Newton’s second law states for every action theirs a equal and opposite reaction, in this case the action would be the parachute will
Open and the reaction to that is that you’re body gets thrown and then finally stabilized. Hope this helped !
Hi there!
Recall the equation for weight.
![W = Mg](https://tex.z-dn.net/?f=W%20%3D%20Mg)
W = Weight (N)
M = Mass (kg)
g = acceleration due to gravity (m/s²)
The weight of an object depends upon its MASS and the strength of the GRAVITATIONAL field. We can solve for weight:
![W = 1360 * 9.8 = \boxed{13,328 N}](https://tex.z-dn.net/?f=W%20%3D%201360%20%2A%209.8%20%3D%20%5Cboxed%7B13%2C328%20N%7D)
Answer: the volleyball
Explanation:
The volleyball, in contrast, is full of air, which contains fewer, more widely spaced particles of matter. In other words, the matter inside the bowling ball is denser than the matter inside the volleyball. A bowling ball is denser than a volleyball.
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>
![y_{f}=0m](https://tex.z-dn.net/?f=y_%7Bf%7D%3D0m)
![v_{iD}=0m/s](https://tex.z-dn.net/?f=v_%7BiD%7D%3D0m%2Fs)
, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so ![t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s](https://tex.z-dn.net/?f=t_%7BE%7D%3Dt_%7BD%7D-t_%7BT%7D%3D7.82s-5.4s%3D2.42s)
a) Speed of the Moon: 1025 m/s
The speed of the moon is equal to the ratio between the circumference of its orbit and the orbital period:
![v=\frac{2 \pi r}{T}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%20%5Cpi%20r%7D%7BT%7D)
where
is the radius of the orbit of the Moon
is the orbital period
Substituting into the formula, we find
![v=\frac{2 \pi (3.85\cdot 10^8 m)}{2.36\cdot 10^6 s)}=1025 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%20%5Cpi%20%283.85%5Ccdot%2010%5E8%20m%29%7D%7B2.36%5Ccdot%2010%5E6%20s%29%7D%3D1025%20m%2Fs)
b) Centripetal force: ![2.0 \cdot 10^{20} N](https://tex.z-dn.net/?f=2.0%20%5Ccdot%2010%5E%7B20%7D%20N)
The centripetal force acting on the Moon is given by:
![F=m\frac{v^2}{r}](https://tex.z-dn.net/?f=F%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
is the mass of the Moon
is its orbital speed
is the radius of the orbit
Substituting into the formula, we find
![F=(7.35 \cdot 10^{22} kg) \frac{(1025 m/s)^2}{3.85\cdot 10^8 m}=2.0 \cdot 10^{20} N](https://tex.z-dn.net/?f=F%3D%287.35%20%5Ccdot%2010%5E%7B22%7D%20kg%29%20%5Cfrac%7B%281025%20m%2Fs%29%5E2%7D%7B3.85%5Ccdot%2010%5E8%20m%7D%3D2.0%20%5Ccdot%2010%5E%7B20%7D%20N)
c) Gravitational force
The only relevant force that acts on the Moon, and that keeps the Moon in circular motion around the Earth, is the gravitational force exerted by the Earth on the Moon. In fact, this force "pulls" the Moon towards the Earth, so towards the centre of the orbit of the Moon, therefore it acts as source of centripetal force for the Moon.