Answer:
The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s
Explanation:
From Newton's second law, F = mg and also from coulomb's law F= Eq
Dividing both equations by mass;
F/m = Eq/m = mg/m, then
g = Eq/m --------equation 1
Again, in a projectile motion, the time of flight (T) is given as
T = (2usinθ/g) ---------equation 2
Substitute in the value of g into equation 2
Charge of proton = 1.6 X 10⁻¹⁹ C
Mass of proton = 1.67 X 10⁻²⁷ kg
E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°
Solving for T;
T = 7.83 X10⁻⁷ s
It has both magnitude and direction
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Explanation:
There are three forces on the bicycle:
Reaction force Rp pushing up at P,
Reaction force Rq pushing up at Q,
Weight force mg pulling down at O.
There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.
Sum of the forces in the y direction:
Rp + Rq − (15)(9.8) = 0
Rp + Rq − 147 = 0
Sum of the moments at P:
(15)(9.8)(0.30) − Rq(1) = 0
44.1 − Rq = 0
Sum of the moments at Q:
Rp(1) − (15)(9.8)(0.70) = 0
Rp − 102.9 = 0
Sum of the moments at O:
Rp(0.30) − Rq(0.70) = 0
0.3 Rp − 0.7 Rq = 0
Any combination of these equations will work.
