Yes the answer is correct
Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;
Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²
Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
Answer:
3secs
Explanation:
Given the following parameters
height H= 81.3m
Velocity v = 12.4m/s
Required
Time it take to reach the ground
Using the equation of motion
H = ut+1/2gt²
81.3 = 12.4t + 1/2(9.8)t²
81.3 = 12.4t + 4.9t²
4.9t² + 12.4t - 81.3 = 0
Using the general formula to find t
t = -12.4±√12.4²-4(4.9)(-81.3)/2(4.9)
t = -12.4±√153.76+1593.48/2(4.9)
t = -12.4±√1747.24/9.8
t = -12.4+41.8/9.8
t = 29.4/9.8
t = 3secs
Hence it took 3secs to reach the ground
