Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Answer:
small intestine, tongue and large intestine
Explanation:
The GI tract is a series of hollow organs joined in a long, twisting tube from the mouth to the anus. The hollow organs that make up the GI tract are the mouth, esophagus, stomach, small intestine, large intestine, and anus. The liver, pancreas, and gallbladder are the solid organs of the digestive system.
Answer:
a = 6.31 m/s²
Explanation:
a) The acceleration of the car can be found using the Newton's second law:
![W = ma](https://tex.z-dn.net/?f=W%20%3D%20ma)
![mgsin(\theta) = ma](https://tex.z-dn.net/?f=mgsin%28%5Ctheta%29%20%3D%20ma)
<u>Where:</u>
F: is the net force acting on the car
m: is the mass
a: is the acceleration of the car
W: is the weight of the car
θ: is the angle = 40.0°
g: is the gravity = 9.81 m/s²
Hence, the acceleration is:
![a = gsin(\theta) = 9.81 m/s^{2}*sin(40.0) = 6.31 m/s^{2}](https://tex.z-dn.net/?f=a%20%3D%20gsin%28%5Ctheta%29%20%3D%209.81%20m%2Fs%5E%7B2%7D%2Asin%2840.0%29%20%3D%206.31%20m%2Fs%5E%7B2%7D)
Therefore, the acceleration of the car is 6.31 m/s².
I hope it helps you!
Answer:
![a = 0.1137 m/s^2](https://tex.z-dn.net/?f=a%20%3D%200.1137%20m%2Fs%5E2)
Explanation:
Let Vc be the velocity of the car and Vm the velocity of the motorcycle. If we convert their given values, we get:
Vc = 87 km/h * 1000m / 1km * 1h / 3600s = 24.17m/s
Vm = 95 km/h * 1000m / 1km * 1h / 3600s = 26.38m/s
Since their positions are equal after 17s we can stablish that:
![Xc = d + Vc*t = Xm = Vm*t + \frac{a*t^2}{2}](https://tex.z-dn.net/?f=Xc%20%3D%20d%20%2B%20Vc%2At%20%20%3D%20Xm%20%3D%20Vm%2At%20%2B%20%5Cfrac%7Ba%2At%5E2%7D%7B2%7D)
Where d is the initial separation distance of 54m. Solving for a, we get:
Repacing the values:
![a = 0.1137 m/s^2](https://tex.z-dn.net/?f=a%20%3D%200.1137%20m%2Fs%5E2)