Answer:
![a=3125000 m/s^2\\a=3.125*10^6 m/s^2](https://tex.z-dn.net/?f=a%3D3125000%20m%2Fs%5E2%5C%5Ca%3D3.125%2A10%5E6%20m%2Fs%5E2)
Acceleration, in m/s, of such a rock fragment = ![3.125*10^6m/s^2](https://tex.z-dn.net/?f=3.125%2A10%5E6m%2Fs%5E2)
Explanation:
According to Newton's Third Equation of motion
![V_f^2-V_i^2=2as](https://tex.z-dn.net/?f=V_f%5E2-V_i%5E2%3D2as)
Where:
is the final velocity
is the initial velocity
a is the acceleration
s is the distance
In our case:
![V_f=V_{escape}, V_i=0,s=4 m](https://tex.z-dn.net/?f=V_f%3DV_%7Bescape%7D%2C%20%20V_i%3D0%2Cs%3D4%20m)
So Equation will become:
![V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2](https://tex.z-dn.net/?f=V_%7Bescape%7D%5E2-V_i%5E2%3D2as%5C%5CV_%7Bescape%7D%5E2-0%3D2as%5C%5CV_%7Bescape%7D%5E2%3D2as%5C%5Ca%3D%5Cfrac%7BV_%7Bescape%7D%5E2%7D%7B2s%7D%5C%5Ca%3D%5Cfrac%7B%285%2A10%5E3m%29%5E2%7D%7B2%2A4%7D%5C%5Ca%3D3125000%20m%2Fs%5E2%5C%5Ca%3D3.125%2A10%5E6%20m%2Fs%5E2)
Acceleration, in m/s, of such a rock fragment = ![3.125*10^6m/s^2](https://tex.z-dn.net/?f=3.125%2A10%5E6m%2Fs%5E2)
Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s
Explanation:
I hope this helps☺️
Answer: They are in the same group because they have similar chemical properties, but they are in different periods because they have very different atomic numbers.
Explanation: On Edgenuity!!
Answer:
![v_f=8.17\frac{m}{s}](https://tex.z-dn.net/?f=v_f%3D8.17%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
![W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7Bx_f%7D_%7Bx_0%7D%20%7BF%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B14%7D_%7B0%7D%20%28%7B18N-0.530%5Cfrac%7BN%7D%7Bm%7Dx%7D%29%20%5C%2C%20dx%5C%5CW%3D%5B%2818N%29x-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7Bx%5E2%7D%7B2%7D%5D%5E%7B14%7D_%7B0%7D%5C%5CW%3D%2818N%2914m-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B%2814m%29%5E2%7D%7B2%7D-%2818N%290%2B%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B0%5E2%7D%7B2%7D%5C%5CW%3D252N%5Ccdot%20m-52N%5Ccdot%20m%5C%5CW%3D200N%5Ccdot%20m)
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
![W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}](https://tex.z-dn.net/?f=W%3D%5CDelta%20K%5C%5CW%3DK_f-K_i%5C%5CW%3D%5Cfrac%7Bmv_f%5E2%7D%7B2%7D-%5Cfrac%7Bmv_i%5E2%7D%7B2%7D)
The box is initially at rest, so
. Solving for
:
![v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}](https://tex.z-dn.net/?f=v_f%3D%5Csqrt%7B%5Cfrac%7B2W%7D%7Bm%7D%7D%5C%5Cv_f%3D%5Csqrt%7B%5Cfrac%7B2%28200N%5Ccdot%20m%29%7D%7B6kg%7D%7D%5C%5Cv_f%3D%5Csqrt%7B66.67%5Cfrac%7Bm%5E2%7D%7Bs%5E2%7D%7D%5C%5Cv_f%3D8.17%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
![m=5.51Kg](https://tex.z-dn.net/?f=m%3D5.51Kg)
Explanation:
The weight of an object on Earth is given by
, so we can calculate its mass by doing
, which for our values is:
![m=(54N)/(9.8m/s^2)=5.51Kg](https://tex.z-dn.net/?f=m%3D%2854N%29%2F%289.8m%2Fs%5E2%29%3D5.51Kg)
<em>Nothing is being asked</em> about Io but if one wanted to know the weight <em>W'</em> of the watermelon there one just have to do:
![W'=ma=(5.51Kg)(1.81m/s^2)=9.97N](https://tex.z-dn.net/?f=W%27%3Dma%3D%285.51Kg%29%281.81m%2Fs%5E2%29%3D9.97N)