What is another word for cycles per second??

Anvisha [2.4K]

The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

$U_i +K_i = U_f + K_f$

where :

$U_i$ is the initial potential energy

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

$K_f$ is the final kinetic energy

The equation can also be rewritten as follows:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 100 kg is the mass of the rock

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 80$ is the initial height

u = 0 is the initial speed (the rock starts at rest)

$h_f = 20 m$ is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

$v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s$

Learn more about kinetic energy and potential energy here:

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5 0

3 years ago

2225.6 grams =________ ft

AnnyKZ [126]

71606.562467

Is the complete answer

6 0

3 years ago

Which of the following is an example of an object that could have a net force greater than zero acting on it?

mario62 [17]

Net force is basically the force an object has when changing direction, so the answer would be D.

5 0

3 years ago

Read 2 more answers

One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

N - $W_m -w_{table}$ = 0