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oksano4ka [1.4K]
2 years ago
6

A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text

for 15.0\,\text m15.0m15, point, 0, start text, m, end text. How much kinetic energy does the dirt gain
Physics
2 answers:
hodyreva [135]2 years ago
5 0

Answer:

77.0k/M

Explanation:

meriva2 years ago
4 0

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

Kinetic Energy = 300Nm

K. E = 1/2

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What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc
Murljashka [212]

Answer:

The Value is V_{AB} = 2.825V

Explanation:

The explanation is shown on the first  uploaded image

5 0
3 years ago
To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal
tatiyna
Answer:  Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v =  linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
 Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).


5 0
3 years ago
Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
2 years ago
A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table
MA_775_DIABLO [31]

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3


consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m


C 2.2 m

3 0
3 years ago
What is an ellipse?
Rom4ik [11]

Answer:

i think it's C thx correct me if wrong

6 0
2 years ago
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