A land rover drops from rest a vertical distance of 1000 m to the surface of planet in 15 seconds what is the magnitude of the a
1 answer:
You might be interested in
Answer:
12.4 m/s²
Explanation:
L = length of the simple pendulum = 53 cm = 0.53 m
n = Number of full swing cycles = 99.0
t = Total time taken = 128 s
T = Time period of the pendulum
g = magnitude of gravitational acceleration on the planet
Time period of the pendulum is given as
T = 1.3 sec
Time period of the pendulum is also given as
g = 12.4 m/s²
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