The ball will decelerate as it moves upwards.
The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
The given parameters;
- initial velocity of the ball, u = 1.25 m/s
- time of motion of the ball, t = 4.22 s
As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.
Thus, the ball decelerate as it moves upwards.
The acceleration of the ball is calculate as;

<em>at the highest point on the incline plane, the final velocity </em>
<em> is zero</em>

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.
Learn more here:brainly.com/question/23860763
Answer:
The second projectile was 1.41 times faster than the first.
Explanation:
In the ballistic pendulum experiment, the speed (v) of the projectile is given by:
<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>
To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:
(1)
<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>
Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

Therefore, the second projectile was 1.41 times faster than the first.
I hope it helps you!
Answer:
A) B = 0.009185 T
B) Drection is negative y-direction
Explanation:
A) We are given;
Speed(v) = 2.5 x 10^(7) m/s
Acceleration (a) = 2.2 x 10^(13) m/s²
We also know that charge of proton(q) = 1.6 x 10^(-19)
Mass of proton(m) = 1.67 x 10^(-27)
Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;
F = qvBsinθ = ma
Since perpendicular, θ = 90°
And so, sinθ = sin 90 = 1
Thus, qvB = ma
Making B the subject gives;
B = ma/qv
B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))
= 0.009185 T
B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction
To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

= Coulomb's constant
q = Charge
r = Radius
At the same time

The values of variables are the same, then if we replace in a single equation we have this expression,

If we replace the values, we have finally that the charge is,




Therefore the potential energy of the system is 