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Natalija [7]
3 years ago
6

A cutting tool several forces acting on it. One force is F=-axy^2 j , a force in the negative y-direction whose magnitude depend

s on the position of the tool. The constant is a= 2.85 . Consider the displacement of the tool from the origin to the point x=2.40 , y=2.40 .. . Calculate the work done on the tool by F if this displacement is along the straight line y=x that connects these two points.. . Calculate the work done on the tool by F if the tool is first moved out along the x-axis to the point x=2.40, y=0 , and then moved parallel to the y-axis to x=2.40
Physics
1 answer:
liq [111]3 years ago
3 0
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.

The work will be completely defined by

                     (Force) x (distance in the y-direction),

and it won't matter what route the tool follows to get anywhere. 
Only the initial and final y-coordinates matter.

We know that    F = - 2.85 y².  (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.

From  y=0  to  y=2.40  is a distance of  2.40  upward.

Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for  Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.

        Work  =  integral of (F·dy) evaluated from  0  to  2.40

                  =  integral of (-2.85 y² dy) evaluated from  0  to  2.40

                 =  (-2.85) · integral of  (y² dy)  evaluated from  0  to  2.40 .


Now, integral of (y² dy)  =  1/3  y³ .

Evaluated from  0  to  2.40 , it's  (1/3 · 2.40³) - (1/3 · 0³)

                                            =  1/3 · 13.824  =  4.608 .

And the work  =  (-2.85) · the integral

                     =  (-2.85) · (4.608)

                     =      - 13.133  .

-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the  -13.133 is joules.

-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40.  Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up. 

-- It doesn't matter whether the tool goes there along the line  x=y , or
by some other route.  WHATEVER the route is, the work done by ' F ' 
is going to total up to be  -13.133 joules at the end of the day.

As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then.  But that's my answer
and I'm stickin to it.  If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
You might be interested in
20 points and brainliest‼️‼️‼️‼️
Anastasy [175]

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

3 0
3 years ago
A book, that has a mass of 0.5 grams, is pushed across a table with a force of 20 newtons. What is the acceleration of the book?
Bas_tet [7]

Answer:

4\cdot 10^4 m/s^2

Explanation:

The acceleration of an object is given by Newton's second law:

a=\frac{F}{m}

where

F is the net force applied on the object

m is the mass of the object

For the book in the problem, we have:

m=0.5 g =5\cdot 10^{-4} kg is the mass

F=20 N is the force applied

Substituting into the formula, we find the acceleration:

a=\frac{20 N}{5\cdot 10^{-4} kg}=4\cdot 10^4 m/s^2

6 0
2 years ago
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun
AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and  i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561  × 10¹⁹ electrons per second = 1.561  × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

8 0
3 years ago
A block with a mass M = 4.85 kg is resting on a slide that has a curved surface. There is no friction. The speed of the block af
natali 33 [55]

Answer:

The correct option is a

Explanation:

From the question we are told that

   The mass of the block is  m =  4.84 \ kg

    The height of the vertical  drop is h =  19.6 \  m

Generally from the law of energy conservation , the potential energy at the top  of the slide is equal to the kinetic energy at the point after sliding this can be mathematically represented as

        PE  =  KE

i.e     m *  g  *   h   =  \frac{1}{2} *  m *  v^2

=>    gh  =  0.5 v^2

=>   v = \sqrt{\frac{9.8 *  19.6}{0.5 } }

=>    v = 19.6 \  m/s

4 0
2 years ago
Se dispone de dos vasos con agua a 20 grados y queremos calentarlo hasta alcanzar 50 grados. Si el primero contiene 0,5l y el se
BigorU [14]

Answer:

The cup with 0.5L

Explanation:

To know what amount of water you take into account the specific heat of the water. The specific heat of water is:

c_{water}=4186\frac{J}{kg\°C}

Thus, 4186 J of energy are needed to icrease the temperature of 1 kg water in 1°C. Then, more grams of water will need more energy.

You have that one cup has 0.5 L and the other one has 750mL = 0.75L

The second cup of water will need more heat because the amount of water contained in the second cup is greater than in the first cup with 0.5L

4 0
3 years ago
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