"<span>H-C=N:" is the one answer among the choices given in the question that shows the correct dot diagram. The correct option among all the options that are given in the question is the fourth option or option "D". The other choices can be neglected. I hope that this is the answer that has come to your help.</span>
To keep the energy going every day
Transverse waves are always characterized by particle motion being perpendicular to wave motion. A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves.
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:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
= 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!