Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]

Now we can use the first statement to get the first equation:

where:
W₁₋₂ = work from the state 1 to 2.


where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]

![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
The density of an object determines whether it will float or sink in another substance. An object will float if it is less dense than the liquid it is placed in. An object will sink if it is more dense than the liquid it is placed in.
So since the boat has a lower density than the water, it will float.
So the answer is choice B
Integrating the velocity equation, we will see that the position equation is:

<h3>How to get the position equation of the particle?</h3>
Let the velocity of the particle is:

To get the position equation we just need to integrate the above equation:


Then:


Replacing that in our integral we get:


Where C is a constant of integration.
Now we remember that 
Then we have:

To find the value of C, we use the fact that f(0) = 0.

C = -1 / 3
Then the position function is:

Integrating the velocity equation, we will see that the position equation is:

To learn more about motion equations, refer to:
brainly.com/question/19365526
#SPJ4
Answer:
yeah this statement is tru
Explanation:
it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.
A testing instrument that's used to measure electrical signals
in a circuit and display them as waveforms on a screen is called
an oscilloscope.