Answer:
When you throw a ball against a wall or drop it onto the floor, you notice that it bounces back up again. In the same way, when the ball hits the floor and gets squished, it pushes back against the floor to try to make itself round again.
Explanation:
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .
also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)
Answer:
The direction of the magnetic field causing this force is
In the plane of the screen and towards the bottom of the egde
Explanation:
This is by applying Fleming s right hand rule which explains that
When a conductor such as a wire attached to a circuit moves through a magnetic field, an electric current is induced in the wire due to Faraday's law of induction. The current in the wire can have two possible directions. Fleming's right-hand rule gives which direction the current flows.
The right hand is held with the thumb, index finger and middle finger mutually perpendicular to each other (at right angles), as shown in the diagram.[1]
The thumb is pointed in the direction of the motion of the conductor relative to the magnetic field.
The first finger is pointed in the direction of the magnetic field. (north to south)
Then the second finger represents the direction of the induced or generated current within the conductor (from the terminal with lower electric potential to the terminal with higher electric potential, as in a voltage source)
Answer:
For these reasons at 98 mph the path is straighter
Explanation:
To solve this problem we are going to use the kinematic equations, specifically those of projectile launches, let's calculate the distances that the ball travels
X = Vox t
Y = Yo + Voy t - ½ g t²
They tell us that the only parameter that changes is the speed, so the distance to the plate is known
t = Vox / x
We replace
Y- Yo = Voy (Vox / X) - ½ g (Vox / x)²2
Y -Yo = Vo² sinθ cos θ / x - ½ g Vo² sin²θ / x²
Y -Yo = Vo² (sinθ cosθ / x - ½ g sin²θ / x²)
The trajectory will be flatter when Y is as close as possible to Yo, when examining the right side of the equation, the amount in Parentheses is constant and to what they tell us that the angles and the distance the plate does not change.
Consequently, of the above, the only amount changes is the initial speed if it increases the square of the same increases, so that the height Y approaches the height of the shoulder, that is, DY decreases. For these reasons at 98 mph the path is straighter
C.
It is a motion with uniform acceleration, meaning that the acceleration will not change.
The object is thrown upwards with a positive velocity. This shows that the upward direction is positive. The object will decelerate due to gravity at a magnitude of 9.81 m/s2. Therefore, the acceleration is -9.81 m/s2.
Note that even though the velocity of the object is momentarily 0 m/s at maximum height, there is still a constant acceleration.
This allows the object first decelerate upwards, then change direction at max height, and finally accelerate downwards. So in this case, the acceleration is always negative and unchanged.
