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
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω = (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω + αt
α= (ω - ω )/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ = ω t + 1/2αt²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω + αt
t= (ω - ω)/α => (0- 3.333)/-1.25
t= 2.67s
Answer:
Explanation:
The answer is a priest or a moulana
Answer:
W_net = mg + 2mgh/r
Explanation:
The forces applied in this motion of the bowling ball are both gravitational and centripetal forces.
Now, gravitational force is; F_g = mg
While centripetal force is; F_c = mv²/r
Since we want to express the net force in terms of the variables in the statement and we are not given "v", let's find an expression of v with the variables given.
Now, from Newton's equation of motion, at initial velocity of 0, v² = 2gh.
Thus;
F_c = 2mgh/r
Where;
m is ball mass
r is tube radius
h is fall height
Thus, the net force will be;
F_net = F_g + F_c
Now, Net force would be equal to the net weight that will be read on the scale.
Thus;
W_net = F_net = F_g + F_c
W_net = mg + 2mgh/r