<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:
Velocity=14[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.
In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.
![Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3D3%5Bkg%5D%5C%5Ch%3D20%5Bm%5D%5C%5C)
![Ep=3*9.81*20\\Ep=588.6[J]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A20%5C%5CEp%3D588.6%5BJ%5D)
When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.
![Ep=3*9.81*10\\Ep=294.3[m]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A10%5C%5CEp%3D294.3%5Bm%5D)
If we know that potential energy is transformed into kinetic energy, we can find the value of speed.
![Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Ctherefore%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%2A2%7D%7Bm%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B294.3%2A2%7D%7B3%7D%20%7D%20%5C%5C%5C%5Cv%3D14%5Bm%2Fs%5D)
Https://courses.physics.illinois.edu/phys102/sp2013/lectures/lecture2.pdf
Check out upload.
Answer:
58.8J
Explanation:
Given parameters;
Mass of ball = 4kg
Height above the floor = 1.5m
g = 9.8n/kg
Unknown:
Potential energy = ?
Solution:
The potential energy of a body is the energy due to the position of the body.
It is mathematically expressed as:
Potential energy = mass x acceleration due to gravity x height
Potential energy = 4 x 9.8 x 1.5 = 58.8J
Answer:
Tarzan will be moving at 7.4 m/s.
Explanation:
From the question given above, the following data were obtained:
Height (h) of cliff = 2.8 m
Initial velocity (u) = 0 m/s
Final velocity (v) =?
NOTE: Acceleration due to gravity (g) = 9.8 m/s²
Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 2.8)
v² = 0 + 54.88
v² = 54.88
Take the square root of both side
v = √54.88
v = 7.4 m/s
Therefore, Tarzan will be moving at 7.4 m/s at the bottom.
