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NARA [144]
3 years ago
7

An object of mass 30KG is falling in air and experiences a force due to air resistance of 50 newtons determine the net force act

ing on the object end and calculate the acceleration of the object use newtons law pleaes!
Physics
1 answer:
ryzh [129]3 years ago
8 0

Answer:

Assume that g = 10\; \rm N \cdot kg^{-1}. The net force on this object will be 250\; \rm N (downwards.) The acceleration of this object will be approximately 8.3\; \rm m \cdot s^{-2} (also downwards.)

Explanation:

<h3>Net force</h3>

The object is falling towards the ground because of gravity. The size of the gravitational force on this object depends on its mass and the strength of the gravitational field at its location.

Near the surface of the earth, the gravitational field strength is approximately 10\; \rm N \cdot kg^{-1}. In other words, approximately 10\; \rm N of gravitational force acts on each kilogram of mass near the surface of the earth.

The mass of this object is given as m = 30\; \rm kg. Therefore, the size of the gravitational force on it will be:

W = m \cdot g \approx 30 \; \rm kg \cdot 10\; N \cdot kg^{-1} = 300\; \rm N.

Near the surface of the earth, gravitational forces point towards the ground. On the other hand, the direction of air resistance on this object will be opposite to its direction of motion. Since this objects is moving towards the ground, the air resistance on it will be directed in the opposite direction. That's exactly the opposite of the direction of the gravitational force on this object. The net force on this object will be:

300\; \rm N - 50\; \rm N =250\; \rm N.

<h3>Acceleration</h3>

Let a denote the acceleration on this object. Apply Newton's Second Law of motion:

\begin{aligned} a &= \frac{F(\text{net force})}{m} \approx \frac{250\; \rm N}{30\; \rm kg} \approx 8.3\; \rm m \cdot s^{-2}\end{aligned}.

Note that the acceleration of this object and the net force on it should be in the same direction.

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PLEASE HELP
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:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

5 0
3 years ago
A rock at the top of a 20 meter tall hill.The rock has a mass of 10 kg. How much potentail   energy does it have?
ch4aika [34]
P.E.=mgh
= 10x10x20
=2000J
6 0
3 years ago
A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc
ki77a [65]
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


7 0
3 years ago
A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?
madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy =\frac{1}{2} mv^2, where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0
2 years ago
Which of the following is NOT a benefit of cool down activities?
quester [9]
B. Maximal lightheadedness
5 0
3 years ago
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