Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
![\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%2Bmg%3D%5Cfrac%7Bmv%5E2%7D%7BL%7D_%7B%7D%20%5C%5C%20v%5E2%3D%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%7D%20%5Cend%7Bgathered%7D)
For the maximum velocity of the ball at the top of the vertical circular motion,
![v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T_%7B%5Cmax%20%7D%2Bmg%29%7D%7Bm%7D%7D)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%287.5_%7B%7D%2B0.31%5Ctimes9.8%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%2810.538%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B17.34%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D4.16%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
Answer:
4 m/s
Explanation:
KE=1/2mv^2
720=1/2(90)
720=45v^2
divide by 45
16= v^2
over the square root
4=v
v= 4m/s
Answer:
blah balh blah blah blah 4 time s for 6 then add 3
Answer:
The speed of the electron is v = 1.01 x 10¹⁵ m/s
Explanation:
Given data,
The mass of the electron, m = 9.11 x 10⁻³¹ kg
The radius of the circle, r = 2.00 x 10⁻² m
The force acting on electron, F = 4.60 x 10⁻¹⁴ N
The speed of the electron, v = ?
The centripetal force of the electron is given by
F = mv² / r
∴ v² = Fr/m
v =√(Fr/m)
Substituting the given values in the above equation,
v =√( 4.60 x 10⁻¹⁴ x 2.00 x 10⁻² / 9.11 x 10⁻³¹ )
v = 1.01 x 10¹⁵ m/s
Hence, the speed of the electron is v = 1.01 x 10¹⁵ m/s
To solve this problem it is necessary to apply the kinematic equations of motion (vertical in this case) as well as the momentum conservation equations.
For conservation of the moment we have to

Where
Mass of each object
Initial velocity of arrow
= Final Velocity
Time in flight for arrow-apple combination is

Where,
h = Max height
g = Gravitational acceleration
Now after the impact arrow-apple combination have horizontal velocity V and it is

From the previous definition we have that the value of time would be,

Assuming the son's height is 1.85m, then we should


Applying again the conservation equation we can obtain the value of the apple mass as:




Therefore the mass of the apple was 140.7g