What is free nitrogen
2 answers:
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Answer:
The rock must leave the cliff at a velocity of 28.2 m/s
Explanation:
The position vector of the rock at a time t can be calculated using the following equation:
r = (x0 + v0x · t, y0 + 1/2 · g · t²)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.
When the rock reaches the ground, the position vector will be (see r1 in the figure):
r1 = (90 m, -50 m)
Then, using the equation of the vector position written above:
90 m = x0 + v0x · t
-50 m = y0 + 1/2 · g · t²
Since x0 and y0 = 0:
90 m = v0x · t
-50 m = 1/2 · g · t²
Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:
-50 m = 1/2 · g · t²
-50 m = -1/2 · 9.81 m/s² · t²
-50 m / -1/2 · 9.81 m/s² = t²
t = 3.19 s
Now, using the equation of the x-component of r1:
90 m = v0x · t
90 m = v0x · 3.19 s
v0x = 90 m / 3.19 s
v0x = 28.2 m/s
Answer:
8.79*10^6 rad/s
Explanation:
To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:
(1)
r: radius of the trajectory
m: mass of the electron = 9.1*10^-31 kg
v: speed of the electron = 1.0*10^6 m/s
q: charge of the electron = 1.6*10^-19 C
B: magnitude of the magnetic field = 5.0*10^-5 T
You use the fact that the angular frequency in a circular motion is given by:
Then, you solve the equation (1) in order to obtain v/r:
Finally, you replace the values of the parameters:
hence, the angular frequency is 8.79*10^6 rad/s
The frequency is: