Answer:
Vertical velocity ![V_y = 10.89 \ m/s](https://tex.z-dn.net/?f=V_y%20%3D%2010.89%20%5C%20m%2Fs)
Horizontal velocity ![V_x = 15.57 \ m/s](https://tex.z-dn.net/?f=V_x%20%3D%2015.57%20%5C%20m%2Fs)
Explanation:
If a 1 -kg is thrown vertically with a velocity of 19 m/s at an angle of 35 °C from a height 1.94 m
The vertical and horizontal component can be resolved as:
![V_y = Vsin \theta \\ \\ \\\\V_x = Vcos \theta](https://tex.z-dn.net/?f=V_y%20%3D%20Vsin%20%5Ctheta%20%5C%5C%20%5C%5C%20%5C%5C%5C%5CV_x%20%3D%20Vcos%20%5Ctheta)
For Vertical component :
![V_y = V sin \theta \\\\V_y = 19 * sin 35 \\ \\ V_y = 19* 0..5736 \\ \\ V_y = 10.89 \ m/s](https://tex.z-dn.net/?f=V_y%20%3D%20V%20sin%20%5Ctheta%20%5C%5C%5C%5CV_y%20%20%3D%2019%20%2A%20sin%2035%20%5C%5C%20%5C%5C%20%20%20V_y%20%3D%2019%2A%200..5736%20%5C%5C%20%5C%5C%20V_y%20%3D%2010.89%20%5C%20m%2Fs)
For horizontal velocity
![V_x = V cos \theta \\\\V_x = 19 * cos 35 \\ \\ V_x = 19* 0.8192 \\ \\ V_x = 15.57 \ m/s](https://tex.z-dn.net/?f=V_x%20%3D%20V%20cos%20%5Ctheta%20%5C%5C%5C%5CV_x%20%20%3D%2019%20%2A%20cos%2035%20%5C%5C%20%5C%5C%20%20%20V_x%20%3D%2019%2A%200.8192%20%5C%5C%20%5C%5C%20V_x%20%3D%2015.57%20%5C%20m%2Fs)
Answer:
A- The ball has both kinetic and potential energy,
Explanation:
kinetic energy by virtue of its motion.
potential energy by virtue of its position. (It could roll off the edge of the tabel and convert gravity potential energy to kinetic energy)
Newton's second law on motion states that a<span>cceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).
So to put it in super simple terms, p</span><span>ushing or pulling an object produces acceleration, a change in the speed of motion.</span>
<span>The momentum of the second ball will depend on the acceleration of the first moving ball.</span>
Answer:
6.02 s
Explanation:
We can write the position of the stock car as:
![x_1 = \frac{1}{2}a_1 t^2](https://tex.z-dn.net/?f=x_1%20%3D%20%5Cfrac%7B1%7D%7B2%7Da_1%20t%5E2)
where
is the acceleration of the stock car.
The sport car instead starts its motion only 1.3 s afterwards, therefore its position at time t can be written as
![x_2 = \frac{1}{2}a_2 (t-1.3)^2](https://tex.z-dn.net/?f=x_2%20%3D%20%5Cfrac%7B1%7D%7B2%7Da_2%20%28t-1.3%29%5E2)
where
is the acceleration of the sport car
(we can verify indeed that when t = 1.3 s,
).
The sport car reaches the stock car when the two positions are equal:
![x_1 = x_2 \\\frac{1}{2}a_1 t^2 = \frac{1}{2}a_2 (t-1.3)^2](https://tex.z-dn.net/?f=x_1%20%3D%20x_2%20%5C%5C%5Cfrac%7B1%7D%7B2%7Da_1%20t%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Da_2%20%28t-1.3%29%5E2)
Rewriting the equation,
![a_1 t^2 = a_2 (t-1.3)^2\\3.2t^2 = 5.2(t-1.3)^2 = 5.2t^2-13.5t+8.8\\2t^2-13.5t+8.8 =0](https://tex.z-dn.net/?f=a_1%20t%5E2%20%3D%20a_2%20%28t-1.3%29%5E2%5C%5C3.2t%5E2%20%3D%205.2%28t-1.3%29%5E2%20%3D%205.2t%5E2-13.5t%2B8.8%5C%5C2t%5E2-13.5t%2B8.8%20%3D0)
This is a second-order equation with two solutions:
t = 0.73 s
t = 6.02 s
We discard the first solution since we are only interested in the times > 1.3 s, therefore the sport car overcomes the stock car after
6.02 seconds.