We know that in a collision the momentum is conserved, that is:
![\vec{p}_0=\vec{p}_f](https://tex.z-dn.net/?f=%5Cvec%7Bp%7D_0%3D%5Cvec%7Bp%7D_f)
Since this is vector equation we can divide it in two scalar equations, one for x and for y. Then we have:
![\begin{gathered} p_{0x}=p_{fx} \\ \text{and} \\ p_{0y}=p_{fy} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20p_%7B0x%7D%3Dp_%7Bfx%7D%20%5C%5C%20%5Ctext%7Band%7D%20%5C%5C%20p_%7B0y%7D%3Dp_%7Bfy%7D%20%5Cend%7Bgathered%7D)
Then we have for the x direction:
![\begin{gathered} 5.4m=m(2.6)\cos 36.9+1.26mv_o\cos \phi \\ 5.4=2.6\cos 36.9+1.26v_o\cos \phi \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%205.4m%3Dm%282.6%29%5Ccos%2036.9%2B1.26mv_o%5Ccos%20%5Cphi%20%5C%5C%205.4%3D2.6%5Ccos%2036.9%2B1.26v_o%5Ccos%20%5Cphi%20%5Cend%7Bgathered%7D)
and for the y direction:
![\begin{gathered} 0=-m2.6\sin 36.9+1.6mv_o\sin \phi \\ 2.6\sin 36.9=1.6v_o\sin \phi \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%200%3D-m2.6%5Csin%2036.9%2B1.6mv_o%5Csin%20%5Cphi%20%5C%5C%202.6%5Csin%2036.9%3D1.6v_o%5Csin%20%5Cphi%20%5Cend%7Bgathered%7D)
Hence, we have the system of equations:
![\begin{gathered} 5.4=2.6\cos 36.9+1.26v_o\cos \phi \\ 2.6\sin 36.9=1.6v_o\sin \phi \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%205.4%3D2.6%5Ccos%2036.9%2B1.26v_o%5Ccos%20%5Cphi%20%5C%5C%202.6%5Csin%2036.9%3D1.6v_o%5Csin%20%5Cphi%20%5Cend%7Bgathered%7D)
From the second equation we have:
![v_o=\frac{2.6\sin 36.9}{1.6\sin \phi}](https://tex.z-dn.net/?f=v_o%3D%5Cfrac%7B2.6%5Csin%2036.9%7D%7B1.6%5Csin%20%5Cphi%7D)
Plugging this in the first equation:
![\begin{gathered} 5.4=2.6\cos 36.9+1.26(\frac{2.6\sin36.9}{1.6\sin\phi})\cos \phi \\ 1.26(\frac{2.6\sin36.9}{1.6})\tan \phi=5.4-2.6\cos 36.9 \\ \tan \phi=\frac{1.6(5.4-2.6\cos 36.9)}{(1.26)(2.6\sin 36.9)} \\ \phi=\tan ^{-1}(\frac{1.6(5.4-2.6\cos36.9)}{(1.26)(2.6\sin36.9)}) \\ \phi=69.69 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%205.4%3D2.6%5Ccos%2036.9%2B1.26%28%5Cfrac%7B2.6%5Csin36.9%7D%7B1.6%5Csin%5Cphi%7D%29%5Ccos%20%5Cphi%20%5C%5C%201.26%28%5Cfrac%7B2.6%5Csin36.9%7D%7B1.6%7D%29%5Ctan%20%5Cphi%3D5.4-2.6%5Ccos%2036.9%20%5C%5C%20%5Ctan%20%5Cphi%3D%5Cfrac%7B1.6%285.4-2.6%5Ccos%2036.9%29%7D%7B%281.26%29%282.6%5Csin%2036.9%29%7D%20%5C%5C%20%5Cphi%3D%5Ctan%20%5E%7B-1%7D%28%5Cfrac%7B1.6%285.4-2.6%5Ccos36.9%29%7D%7B%281.26%29%282.6%5Csin36.9%29%7D%29%20%5C%5C%20%5Cphi%3D69.69%20%5Cend%7Bgathered%7D)
Now that we know the value of the angle we plug it in the expression for the velocity, then we have:
![\begin{gathered} v_o=\frac{2.6\sin 36.9}{1.6\sin 69.69} \\ v_0=1.04 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_o%3D%5Cfrac%7B2.6%5Csin%2036.9%7D%7B1.6%5Csin%2069.69%7D%20%5C%5C%20v_0%3D1.04%20%5Cend%7Bgathered%7D)
Therefore, the magnitude of the final speed of the orange ball is 1.04 m/s and the direction is 69.69°
Answer:
213.5 kJ
Explanation:
Mechanical energy= kinetic energy + potential energy= KE+PE=
where m is the mass of rock climber, v is the velocity, g is acceleration due to gravity and h is the height.
By substitution where 68 Kg is m, 320 m is h, 1.5 m/s is v and 9.81 is g then
Mechanical energy=![68(0.5\times 1.5^{2}+9.81\times 320)=213542.1 J\approx 213.5 kJ](https://tex.z-dn.net/?f=68%280.5%5Ctimes%201.5%5E%7B2%7D%2B9.81%5Ctimes%20320%29%3D213542.1%20J%5Capprox%20213.5%20kJ)
Answer:
the sheets approach while the object is near
Explanation:
An electroscope is an apparatus that has two metal sheets attached, when these sheets have a charge and distribute evenly between them and the sheets repel.
When I approach an object charged with a counter (negative) charge, part of the charge of the electroscope moves near the charged external object, to neutralize the electric field, so as the charge on one of the plates decreases the electroscope has approached , as the objects are not touched the system remains in this configuration while the object is close. When the object is released, the electric field it creates disappears, so the positive charges repel inside the electroscope and the sheets repel to the initial position.
In short, the sheets approach while the object is near
Answer : distance, d = 463.7 m
Explanation :
According to the given condition it can be assumed that the first echo would be heard from closest cliff. The second echo is from farther cliff and the third echo is from the reflection between the two cliffs.
Let the distance between the first cliff and the point of firing is x and y is the distance between the second cliff and the point of firing.
Then the first echo will travel 2x distance, second will travel 2y distance and third will travel 2x +2y.
So, using above data :
![2y=v\times t_3](https://tex.z-dn.net/?f=2y%3Dv%5Ctimes%20t_3)
and
![2x=v(t_2+t_3)](https://tex.z-dn.net/?f=2x%3Dv%28t_2%2Bt_3%29)
On solving :
![y=\dfrac{343\times0.827}{2}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B343%5Ctimes0.827%7D%7B2%7D)
![y=141.8\ m](https://tex.z-dn.net/?f=y%3D141.8%5C%20m)
![x=\frac{343\times(1.05+0.827)}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B343%5Ctimes%281.05%2B0.827%29%7D%7B2%7D)
![x=321.9\ m](https://tex.z-dn.net/?f=x%3D321.9%5C%20m)
x = 321.9 m and y = 141.8 m
Hence, total distance between two cliffs is d = 321.9 m + 141.8 m = 463.7 m