<span>51 degrees.
Since we're ignoring friction, we have to have a banking angle such that the normal force is exactly perpendicular to the banked curve. Since this problem says "ignore friction", if the bank angle is too shallow, the bobsled would slide outwards if the banking angle is too shallow and would fall inwards if the banking angle is too steep. So we have to exactly match the calculated centripetal acceleration.
The equation for centripetal acceleration is:
F = mv^2/r
I'll assume a mass of 1 kg to keep the math simple. Any mass could be used and the direction vectors would be the same except their magnitude would differ. So
F = 1 kg * (35 m/s)^2/100 m
F = 1225 kg*m^2/s^2 / 100 m
F = 12.25 kg*m/s^2
The local gravitational acceleration is 9.8 m/s^2, so the sum of those vectors will have a length of sqrt(12.25^2 + 9.8^2) and an angle of atan(9.8/12.25) below the horizon. The magnitude of the vector doesn't matter, merely the angle which is:
atan(9.8/12.25) = atan(0.8) = 38.65980825 degrees.
The banking angle needs to be perpendicular to the force vectors. So
90 - 38.65980825 = 51.34019175 degrees.
Rounding to 2 significant figures gives a bank angle of 51 degrees.</span>
Answer:
So, the radius of fifth Bohr orbital of hydrogen is 1. 3225 nm.
Explanation:
pls mark me brainless hope this helps loves x!
I think it is -3.99 x 102 j
Answer:
Light provides brightness to see and also light carries energy
Kinetic energy = momentum^2 / 2 x mass
Mass = (momentum^2/ Kinetic energy) / 2
Mass = (25^2 / 275) / 2
Mass = 1.136 kg.
momentum = mass x velocity
velocity = mass / momentum
velocity = 1.136 / 25
velocity = 0.04544 m/s
