A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
Answer:
5,865,696,000,000
Explanation:
31,536,000 seconds in a year. 31,536,000×186,000=5,865,696,000,000
To solve this problem it is necessary to apply the concepts related to energy conservation. Therefore, the work done will initially be equivalent to the change in kinematic energy. And this kinematic energy will be equivalent to the sum of energy (work) carried out by force and friction. In this way we have to
![E = \frac{1}{2} m (v_f^2-v_i^2)](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%20%28v_f%5E2-v_i%5E2%29)
![E = \frac{1}{2} (18.8kg)((2.9m/s)^2-(1.5m/s)^2)](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2818.8kg%29%28%282.9m%2Fs%29%5E2-%281.5m%2Fs%29%5E2%29)
![E = 57.904J](https://tex.z-dn.net/?f=E%20%3D%2057.904J)
Now,
![E = \Delta W](https://tex.z-dn.net/?f=E%20%3D%20%5CDelta%20W)
![E = W_{Force}+W_{Friction}](https://tex.z-dn.net/?f=E%20%3D%20W_%7BForce%7D%2BW_%7BFriction%7D)
For each one we have
![W_{Force} = F dcos(\theta)](https://tex.z-dn.net/?f=W_%7BForce%7D%20%3D%20F%20dcos%28%5Ctheta%29)
![W_{Force} = (160)(6.20*cos(26))](https://tex.z-dn.net/?f=W_%7BForce%7D%20%3D%20%28160%29%286.20%2Acos%2826%29%29)
![W_{Force} = 891.604J](https://tex.z-dn.net/?f=W_%7BForce%7D%20%3D%20891.604J)
Then at the equation of equilibrium of energy we have,
![57.904J = 891.604J+W_{Friction}](https://tex.z-dn.net/?f=57.904J%20%3D%20891.604J%2BW_%7BFriction%7D)
![W_{Friction} = 57.904-891.604](https://tex.z-dn.net/?f=W_%7BFriction%7D%20%3D%2057.904-891.604)
![W_{Friction} = -833.7J](https://tex.z-dn.net/?f=W_%7BFriction%7D%20%3D%20-833.7J)
Therefore the work done by friction on the chair is -833.7J
Answer:
See the explanation below
Explanation:
In the attached image we can see the free body diagram and forces acting on the book.
Force F, which acts to the right, the friction force which acts in the opposite direction to the movement, i.e. to the left.
The force exerted by the weight of the book that is equal to the product of the mass of the book by the gravity of the book. The normal force that is equal in magnitude to the weight of the book but in the opposite direction.