Answer:
Explanation:
Given
Object is thrown with a velocity of 
Acceleration due to gravity is -g (i.e. acting downward)
Vertical distance traveled by object is given by
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
at maximum height final velocity is zero
time taken to reach maximum height
using
v=u+at
0=9-gt
If a ship will be sailing through warm and cold water, people think about making it less dense than the warmest water as they load the ship with cargo. I think you forgot to give the options along with the question. I hope that this is the answer that has actually come to your desired help.
Answer:
a = 2 [m/s²]
Explanation:
To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.
We can use the following equation.
where:
Vf = final velocity = 11 [m/s]
Vo = initial velocity = 5 [m/s]
a = acceleration [m/s²]
t = time = 3 [s]
![11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]](https://tex.z-dn.net/?f=11%20%3D%205%20%2B%20a%2A3%5C%5C6%3D3%2Aa%5C%5Ca%3D%202%5Bm%2Fs%5E%7B2%7D%20%5D)
Answer:
The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.
Explanation:
Answer:
296 N
Explanation:
Draw a free body diagram. The box has two forces on it: tension up and weight down.
Apply Newton's second law:
∑F = ma
T − mg = ma
T = m (g + a)
Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:
T = (20 kg) (9.8 m/s² + 5 m/s²)
T = 296 N
