Answer:
Hiii how are you <u>doing?</u><u>?</u><u>I </u><u>don't</u><u> </u><u>understand</u><u> </u><u>that</u>
Answer:
If we’re talking about objects on the Earth, the gravitational potential energy is given by:
Explanation:
PEg=mgh
so the energy is proportional to the mass ( m ), but also to the strength of the gravitational field ( g ), and the height ( h ) to which the mass is lifted.
1)

<Solve using the formula which is:
Mass=Density×Volume

2)
3)

(Length × Width × Height)


(Answer=7.5)
Answer:
16.1 m/s
Explanation:
We can solve the problem by using the law of conservation of energy.
At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

where m = 0.15 kg is the mass of the block and v is its speed.
Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

<span>Mass of the block m = 3.3kg
Angle of the slide = 30 degrees
Distance the block slides s = 2.10 m
Time taken to slide t = 1.6 s
Initially in rest condition so initial velocity u = 0.
We have an equation for distance s = (u x t) + (1/2) x (a t^2)
s = (0 x t) + (1/2) x (a x (1.6) ^2) => 2.10 = (1/2) x (a x2.56)
2.56a = 4.20 => a = 1.64
So the magnitude of the Acceleration a = 1.64 m/s^2</span>