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Setler [38]
3 years ago
12

with what speed does a freely falling object dropped from a height of 88.2m hit the ground? how long does it take for it hit the

ground?
Physics
1 answer:
maria [59]3 years ago
6 0

Answer:

Vf = 41.6 [m/s]

Explanation:

To solve this problem we must use the equations of kinematics.

Vf² = Vo² + (2*g*y)

where:

Vf =  final velocity [m/s]

Vo = initial velocity = 0

g = gravity acceleration = 9.81 [m/s²]

y = height = 88.2 [m]

Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.

Vf² = Vo² + (2*g*y)

Vf² = 0 + (2*9.81*88.2)

Vf = (1730.48)^0.5

Vf = 41.6 [m/s]

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

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