No artificial ingredients
i only know that because my mom was vegan for two years
Hey there!
Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.
0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s
–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s
So, the cars are traveling at -0.33 m/s in the direction of the second car.
Hope this helps
<em>Tobey</em>
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s
Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
The change in gravitational potential energy due to change in position must be the change in it's kinetic energy as the system is isolated! so find out the potential energies of the two different points!
<span>PE=−[G<span>M1</span><span>M2</span>]÷R
</span><span>
Potential energy of a particle due to mass A is not affected by presence of any other mass B !</span>
