The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
1 answer:
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Answer:
velocity = 7.26 * 10^6 m/sec
Explanation:
The rule that is used to solve this problem is shown in the attached image.
The variables are as follows:
k = 8.99 * 10^9 Nm^2 / C^2
e is the electron charge = -1.6 * 10^-19 C
q is the charge given = 1 * 10^-9 C
m is the mass of the electron = 9.11 * 10^-31
r1 is the radius of starting point = 3 cm = 0.03 m
r2 is the radius of the sphere = 2 cm = 0.02 m
Substitute with the givens in the equation to get the value of the velocity
Hope this helps :)
velocity = 7.26 * 10^6 m/sec
Explanation:
The rule that is used to solve this problem is shown in the attached image.
The variables are as follows:
k = 8.99 * 10^9 Nm^2 / C^2
e is the electron charge = -1.6 * 10^-19 C
q is the charge given = 1 * 10^-9 C
m is the mass of the electron = 9.11 * 10^-31
r1 is the radius of starting point = 3 cm = 0.03 m
r2 is the radius of the sphere = 2 cm = 0.02 m
Substitute with the givens in the equation to get the value of the velocity
Hope this helps :)
(a) 18.9 m/s
The motion of the stone consists of two independent motions:
- A horizontal motion at constant speed
- A vertical motion with constant acceleration () downward
We can calculate the components of the initial velocity of the stone as it is launched from the ground:
The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.
When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore
(b) 22.2 m/s
We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by
(1)
where
is the initial vertical velocity
is the vertical velocity at height h
is the acceleration due to gravity (negative because it is downward)
At the top of the parabolic path, , so we can use the equation to find the maximum height
So, at half of the maximum height,
And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:
And so, the speed of the stone at half of the maximum height is
(c) 17.4% faster
We said that the speed at the top of the trajectory (part a) is
while the speed at half of the maximum height (part b) is
So the difference is
And so, in percentage,
So, the stone in part (b) is moving 17.4% faster than in part (a).