Weight = mg Mass, m = 8kg, g ≈ 9.8 m/s² on Earth
W = 8*9.8
W ≈ 78.4 N
Answer:
As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don't actually travel along with the wave. Only the energy of the wave travels through the medium.
Answer:
2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂
2Na + MgCl₂ → 2NaCl + Mg
Explanation:
A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation
Among the given chemical reactions, we have;
2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂
In the above reaction;
The number of phosphorus, P, on either side of the equation = 2
The number of bromine atoms, Br, on either side of the equation = 6
The number of chlorine atoms, Cl, on either side of the equation = 6
Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced
The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg
In the above reaction, there are two sodium atoms, Na, one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced
Explanation:
(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.
(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.
First we must find the rocket speed when the engines stop:
![v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}](https://tex.z-dn.net/?f=v_f%5E2%3Dv_0%5E2%2B2ay_1%5C%5Cv_f%5E2%3D%2852%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%2B2%281%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%28160m%29%5C%5Cv_f%5E2%3D3024%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%5C%5Cv_f%3D%5Csqrt%7B3024%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%3D54.99%5Cfrac%7Bm%7D%7Bs%7D)
This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:
![v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m](https://tex.z-dn.net/?f=v_f%5E2%3Dv_0%5E2%2B2gy_%7B2%7D%5C%5Cy_%7B2%7D%3D%5Cfrac%7Bv_f%5E2-v_0%5E2%7D%7B2g%7D%5C%5Cy_%7B2%7D%3D%5Cfrac%7B0%5E2-%2854.99%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%7D%7B2%28-9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%7D%3D154.28m)
So, the maximum height reached by the rocket is:
![h=y_1+y_2\\h=160m+154.28m=314.28m](https://tex.z-dn.net/?f=h%3Dy_1%2By_2%5C%5Ch%3D160m%2B154.28m%3D314.28m)
(c) In the first part we have:
![v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s](https://tex.z-dn.net/?f=v_f%3Dv_0%2Bat_1%5C%5Ct_1%3D%5Cfrac%7Bv_f-v_0%7D%7Ba%7D%5C%5Ct_1%3D%5Cfrac%7B54.99%5Cfrac%7Bm%7D%7Bs%7D-52%5Cfrac%7Bm%7D%7Bs%7D%7D%7B1%5Cfrac%7Bm%7D%7Bs%5E2%7D%7D%5C%5Ct_1%3D2.99s)
And in the second part:
![t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s](https://tex.z-dn.net/?f=t_2%3D%5Cfrac%7Bv_f-v_0%7D%7Bg%7D%5C%5Ct_2%3D%5Cfrac%7B0-54.99%5Cfrac%7Bm%7D%7Bs%7D%7D%7B-9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%7D%5C%5Ct_2%3D5.61s)
So, the time it takes to reach the maximum height is:
![t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s](https://tex.z-dn.net/?f=t_3%3Dt_1%2Bt_2%5C%5Ct_3%3D2.99s%2B5.61s%3D8.60s)
(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:
![v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}](https://tex.z-dn.net/?f=v_f%5E2%3Dv_0%5E2%2B2ay%5C%5Cv_f%5E2%3D0%5E2%2B2%289.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%29%28314.28m%29%5C%5Cv_f%3D%5Csqrt%7B6159.888%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%3D78.48%5Cfrac%7Bm%7D%7Bs%7D)
![t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s](https://tex.z-dn.net/?f=t_4%3D%5Cfrac%7Bv_f-v_0%7D%7Bg%7D%5C%5Ct_4%3D%5Cfrac%7B78.48%5Cfrac%7Bm%7D%7Bs%7D-0%7D%7B9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%7D%5C%5Ct_4%3D8.01s)
So, the total time is:
![t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s](https://tex.z-dn.net/?f=t%3Dt_3%2Bt_4%5C%5Ct%3D8.60s%2B8.01s%5C%5Ct%3D16.61s)
Answer:
9] V = D ÷ T
Take any distance value from the graph and its relevant time.
V = 4 ÷ 2
V = 2 m/s
[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]
10] Take the distance value and its time for the highest peak of B.
V = 20 ÷ 2
V = 10 m/s