Answer:The gravitational force on the ball due to earth is exactly the same as the gravitational force on earth due to the ball
Explanation:
The gravitational force exist in pair. It form action-reaction pair.
Answer:Due to change in direction
Explanation:
Given
Initially train has traveled a 100 km in North and after exchanging some railroad cars, it traveled 100 in south.
The velocity of the train changes as it direction of motion changes. Velocity is the vector quantity which require direction and magnitude for its reperesentation.
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)
An Isotope has the same number of Protons but a different number of neutrons than other atoms of the same element.
In short, an isotope has the same atomic number, but a different molar mass.
Answer:
As per law of conservation of momentum
1 × 10 = 50× v
v = 1/5 m/S
Or
20 cm/s