1) The mass of the continent is 
2) The kinetic energy of the continent is 274.8 J
3) The speed of the jogger must be 2.76 m/s
Explanation:
1)
The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:
The mass of the continent is given by
where:
is its density
is its volume
Substituting, we find the mass:
2)
To find the kinetic energy, we need to convert the speed of the continent into m/s first.
The speed is
v = 1.6 cm/year
And we have:
1.6 cm = 0.016 m
So, the speed is
Now we can find the kinetic energy of the continent, which is given by
where
is the mass
is the speed
Substituting,
3)
The jogger in this part has the same kinetic energy of the continent, so
K = 274.8 J
And its mass is
m = 72 kg
We can write his kinetic energy as
where
v is the speed of the man
And solving the equation for v, we find his speed:
Learn more about kinetic energy:
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Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve 
is 
.
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
Where
, 
- Initial and final position, respectively, measured in meters.
- Force as a function of position, measured in newtons.
Given that 
and the fact that 
when 
, the spring constant (
), measured in newtons per meter, is:
Now, the work function is obtained:
![W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]](https://tex.z-dn.net/?f=W%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%28250%5C%2C%5Cfrac%7BN%7D%7Bm%7D%20%5Cright%29%5Ccdot%20%5B%280.05%5C%2Cm%29%5E%7B2%7D-%280.00%5C%2Cm%29%5E%7B2%7D%5D)
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be . The area of the region enclosed by one loop of the curve is given by the following integral:
![A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta](https://tex.z-dn.net/?f=A%20%3D%20%5Cint%5Climits%5E%7B2%5Cpi%7D_0%20%7B%5Br%28%5Ctheta%29%5D%5E%7B2%7D%7D%20%5C%2C%20d%5Ctheta)
By using trigonometrical identities, the integral is further simplified:
The area of the region enclosed by one loop of the curve is .
Answer:
v=77.62 m/s
Explanation:
Given that
h= - 300 m
speed of the bird ,u= 5 m/s
Lets take Speed of the berry when it hit the ground = v m/s
we know that ,if object is moving upward
v² = u² - 2 g h
u=Initial speed
v=Final speed
h=Height
Now by putting the values
v² = u² - 2 g h
v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)
v² =25 + 20 x 300
v² ==25 + 6000
v² =6025
v=77.62 m/s
Therefore the final speed of the berry will be 77.62 m/s.
