Answer:
(a) The total energy of the object at any point in its motion is 0.0416 J
(b) The amplitude of the motion is 0.0167 m
(c) The maximum speed attained by the object during its motion is 0.577 m/s
Explanation:
Given;
mass of the toy, m = 0.25 kg
force constant of the spring, k = 300 N/m
displacement of the toy, x = 0.012 m
speed of the toy, v = 0.4 m/s
(a) The total energy of the object at any point in its motion
E = ¹/₂mv² + ¹/₂kx²
E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²
E = 0.0416 J
(b) the amplitude of the motion
E = ¹/₂KA²
![A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%7B%5Cfrac%7B2E%7D%7BK%7D%20%7D%20%5C%5C%5C%5CA%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2A0.0416%7D%7B300%7D%20%7D%20%5C%5C%5C%5CA%20%3D%200.0167%20%5C%20m)
(c) the maximum speed attained by the object during its motion
![E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv_%7Bmax%7D%5E2%5C%5C%5C%5Cv_%7Bmax%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2E%7D%7Bm%7D%20%7D%20%5C%5C%5C%5Cv_%7Bmax%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%2A0.0416%7D%7B0.25%7D%20%7D%20%5C%5C%5C%5Cv_%7Bmax%7D%20%3D%200.577%20%5C%20m%2Fs)
A nuclear can hold 2 electrons
Answer:
C. 590 mph
![\vert v_{cj}\vert=589.49\ mph](https://tex.z-dn.net/?f=%5Cvert%20v_%7Bcj%7D%5Cvert%3D589.49%5C%20mph)
Explanation:
Given:
- velocity of jet,
![v_j=500\ mph](https://tex.z-dn.net/?f=v_j%3D500%5C%20mph)
- direction of velocity of jet, east relative to the ground
- velocity of Cessna,
![v_c=150\ mph](https://tex.z-dn.net/?f=v_c%3D150%5C%20mph)
- direction of velocity of Cessna, 60° north of west
Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.
Refer the attached schematic.
So,
![\vec v_j=500\ \hat i\ mph](https://tex.z-dn.net/?f=%5Cvec%20v_j%3D500%5C%20%5Chat%20i%5C%20mph)
&
![\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)](https://tex.z-dn.net/?f=%5Cvec%20v_c%3D150%5Ctimes%20%28%5Ccos120%5C%20%5Chat%20i%2B%5Csin120%5C%20%5Chat%20j%29)
![\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph](https://tex.z-dn.net/?f=%5Cvec%20v_c%3D-75%5C%20%5Chat%20i%2B75%5Csqrt%7B3%7D%5C%20%5Chat%20j%5C%20mph)
Now the vector of relative velocity of Cessna with respect to jet:
![\vec v_{cj}=\vec v_j-\vec v_c](https://tex.z-dn.net/?f=%5Cvec%20v_%7Bcj%7D%3D%5Cvec%20v_j-%5Cvec%20v_c)
![\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )](https://tex.z-dn.net/?f=%5Cvec%20v_%7Bcj%7D%3D500%5C%20%5Chat%20i-%28-75%5C%20%5Chat%20i%2B75%5Csqrt%7B3%7D%5C%20%5Chat%20j%20%29)
![\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph](https://tex.z-dn.net/?f=%5Cvec%20v_%7Bcj%7D%3D575%5C%20%5Chat%20i-75%5Csqrt%7B3%7D%5C%20%5Chat%20j%5C%20mph)
Now the magnitude of this velocity:
![\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}](https://tex.z-dn.net/?f=%5Cvert%20v_%7Bcj%7D%5Cvert%3D%5Csqrt%7B%28575%29%5E2%2B%2875%5Csqrt%7B3%7D%20%29%5E2%7D)
is the relative velocity of Cessna with respect to the jet.
Answer:
Reflected
Explanation:
I do not have much context here, but reflection is what happens when the sun sets on the water. The rays hit the surface of the water and bounce off, known as refelction.
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather