Answer:
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
Explanation:
The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.
Given are
<em> mass of the student 73 kg</em>
<em> height of water glide 11.8 m</em>
<em> work done as -5.5*10³ J</em>
Have to find speed at which the student goes down the glide.
According to<em> Law of Conservation of energy</em>,
K.E =P.E+Work Done
mv²/2=mgh +W
Rearranging the above eqn for v
v = √2(gh+W/m)
Substituting values,
V = 12.48 m/s.
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
The answer for this question is letter "B.Fission releases energy, and its products have greater stability."
Fission and Fusion are both nuclear reactions that when they release energy, they make the nuclei more stable. So among the choices, option B is the most fitting for the definition.
Answer:
770m/s
Explanation:
caculation using one of the newton law of motion
Hello!
a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.
At the top of the circle, we have:
- Force of gravity (downward)
*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.
Therefore, only the force of gravity produces the net centripetal force:

We can simplify and rearrange the equation to solve for 'v'.

Plugging in values:

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)
- Tension force (upward, +)
The sum of these forces produces a centripetal force, upward (+).

Rearranging for 'T":

Plugging in the appropriate values:

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).
We can calculate the initial velocity of the car (u) using the following kinematic equation.
![v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s](https://tex.z-dn.net/?f=v%5E%7B2%7D%20%3D%20u%5E%7B2%7D%20%2B%202as%5C%5C%5C%5Cu%20%3D%20%5Csqrt%5B%5D%7Bv%5E%7B2%7D-2as%7D%20%3D%20%5Csqrt%5B%5D%7B%280m%2Fs%29%5E%7B2%7D-2%28-42.61m%2Fs%5E%7B2%7D%20%29%2810.99m%29%7D%20%3D%2030.60m%2Fs)
A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.
Learn more: brainly.com/question/14851168