Find the equivalent resistance of this

For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener

Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

$v=u+at$

where:

v is the velocity at time t

u is the initial velocity

$a=g=9.8 m/s^2$ is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

$u=0$ (the object starts from rest)

$a=9.8 m/s^2$

Therefore, the velocity after

t = 2 s

is:

$v=0+(9.8)(2)=19.6 m/s$ (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

$KE=\frac{1}{2}(3)(19.6)^2=576 J$

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

$a=g=9.8 m/s^2$ (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

$s=0+\frac{1}{2}(9.8)(2)^2=19.6 m$

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

$KE'=\frac{1}{2}Mv^2$

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

$KE'=\frac{1}{2}(6)(19.6)^2=1152 J$

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0

2 years ago

A cylinder at left with balls evenly spaced throughout the cylinder has an arrow leading to a cylinder at right cylinder with ba

pentagon [3]

The change of state as depicted by the model is condensation.

<h3>What is Condensation?</h3>

The process by which the water vapors get converted into liquid droplets to get settled on a cold surface is known as condensation.

As per the given problem, the cylinder in the left has the balls evenly spaced throughout and the right cylinder has the balls in the form of layers at its bottom.

So, this clearly reflects that the comparing both the balls in both cylinders, the right cylinder consists of the balls undergone through the process of condensation.

This is because,

During condensation, the vapor molecules get converted into liquid.

And in the left cylinder, the balls evenly spaced have more intermolecular distance which designates the gaseous molecules.

While in the right cylinder, the layered form of gas shows that the conversion occurred to some state of molecules, having less intermolecular space.

And liquid has less intermolecular space than gas.

Thus, we can conclude that the change of state as depicted by the model is condensation.

Learn more about the condensation here:

brainly.com/question/15563071

8 0

2 years ago

Read 2 more answers

The planet Mercury takes 0.24 sidereal years to go around the sun. What is the distance from the center of Mercury to the center

lyudmila [28]

Answer:

5.77×10¹⁰ m

Explanation:

From the question,

Applying

Kepler's third law

P² = d³...................... Equation 1

Where P = Planet's period, d = distance between the center of the planet and the sun.

make d the subject of the formula n equation 1

d = $\sqrt[3]{P^{2} }$ .................. Equation 2

Given: P = 0.24 sidereal.

Substitute the value of P into equation 2

d = $\sqrt[3]{0.24^{2} }$

d = 0.386 Au

d = 0.386×1.496×10¹¹

d = 5.77×10¹⁰ m

6 0

3 years ago

A quarterback takes the ball from the line of scrimmage, runs backward for 12.1 yds, then runs sideways parallel to the line of

Over [174]

Answer:

The answer is 35.45 yds

Explanation:

You have to picture this to be able to understand it better (see attachment).

Start at the origin which is when the quarterback (QB) takes the ball. He runs backwards 12.1 yards, runs sideways for 19.8 yards (it doesn´t matter if he runs right or left), then he throws the ball forward 41.5 yards. If you look at the attachment, you can see I drew the path that the football followed. And then connected the dots from the origin and finish. The distance between those two points is the magnitude of the resultant displacement.

In order to calculate it, all you need to do is use the Pythagoream theorem, which says that the square of the hypotenuse equals the sum of the squares of the legs a and b of the triangle rectangle.

$R^{2} = a^{2} + b^{2}$ then solve for R
$R = \sqrt{a^{2}+b^{2} }$

In this case, you know the length of leg a to be 19.8 yards which how much it moves sideways. And then, to get the length of leg b, all you need to do is substract how much it moved backwards from the 41.5yards forward displacement. This results in b leg being 29.4 yards long.

Now you have a triangle with:

a = 19.8 yards
b = 29.4 yards

Substituting this numbers in the equation:

$R = \sqrt{19.8^{2}+29.4^{2} }$
R = 35.45 yards

3 0

2 years ago

Suppose the tourist in question #1 instead threw the rock with an initial velocity of 8.0

Harman [31]

Answer:

Explanation:u=8m/s, a=8m/s*s, t=4s

Using v= u + at

v=8+-8*4

v=-24m/s

Displacement=s

Using s=½(u+v)t

s=½(8-24)4

s=½*-16*4

s=-8*4

s=32m( you can solve without observing the negative sign)

8 0

3 years ago