Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,
![\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\](https://tex.z-dn.net/?f=%5Cint%5Climits%5Es_0%20%7Ba%7D%20%5C%2C%20ds%20%20%3D%20%5Cint%5Climits%5Ev_0%20%7Bv%7D%20%5C%2C%20dv%5C%5C%20%5Cint%5Climits%5Es_0%20%7B6%2B0.02s%7D%20%5C%2C%20ds%20%20%3D%20%5Cint%5Climits%5Ev_0%20%7Bv%7D%20%5C%2C%20dv%5C%5C%206s%20%2B%20%5Cfrac%7B0.02s%5E%7B2%7D%20%7D%7B2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20v%5E%7B2%7D%20%5C%5Cv%20%3D%20%5Csqrt%7B12s%20%2B%200.02s%5E%7B2%7D%20%7D%20.....................1%20%5C%5C%5C%5C%5C%5C)
Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
<span>In the desert environment the chemical weathering of rocks is generally reduced because there is a lack of rain, in which most chemical weathering is caused by. No rain, no chemicals, no chemical weathering.</span>
Answer:
Part a)
A = 0.066 m
Part b)
maximum speed = 0.58 m/s
Explanation:
As we know that angular frequency of spring block system is given as
![\omega = \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
here we know
m = 3.5 kg
k = 270 N/m
now we have
![\omega = \sqrt{\frac{270}{3.5}}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B270%7D%7B3.5%7D%7D)
![\omega = 8.78 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%208.78%20rad%2Fs)
Part a)
Speed of SHM at distance x = 0.020 m from its equilibrium position is given as
![v = \omega \sqrt{A^2 - x^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20%5Csqrt%7BA%5E2%20-%20x%5E2%7D)
![0.55 = 8.78 \sqrt{A^2 - 0.020^2}](https://tex.z-dn.net/?f=0.55%20%3D%208.78%20%5Csqrt%7BA%5E2%20-%200.020%5E2%7D)
![A = 0.066 m](https://tex.z-dn.net/?f=A%20%3D%200.066%20m)
Part b)
Maximum speed of SHM at its mean position is given as
![v_{max} = A\omega](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%20A%5Comega)
![v_{max} = 0.066(8.78) = 0.58 m/s](https://tex.z-dn.net/?f=v_%7Bmax%7D%20%3D%200.066%288.78%29%20%3D%200.58%20m%2Fs)
8 x 10^8 = 800,000,000
In Scientific Notation, your goal is to get your the number you're multiplying by 10 (8 in this case) to be between 0 and 10. Therefore, you would NOT have 80 x 10^7 because 80 is not between 0 and 10.
Answer:
Friction is useful in some cases like walking and cycling ..
but it is unwanted in machines as it create unwanted sounds and heat .,due to which we loss energy
Explanation:
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