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3 years ago

11

A boy pulls a wagon with an applied force of 40 N on frictionless surface. If the mass of the wagon is 13kg, what is the acceler

ation of the wagon?

1 answer:

Gekata [30.6K]3 years ago

7 0

Answer:

3.1 m/s²

Explanation:

Apply Newton's second law:

∑F = ma

40 N = (13 kg) a

a ≈ 3.1 m/s²

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Which atomic model was proposed as a result of j. J. Thomson’s work?

EleoNora [17]

<h2>Answer: The "raising pudding" atomic model</h2>

Explanation:

<u>During the 19th century the accepted atomic model, was Dalton's atomic model</u>, which postulated the atom was an <u>"individible and indestructible mass".</u>

However, at the end of 19th century J.J. Thomson began experimenting with cathode ray tubes and found out that atoms contain small subatomic particles with a negative charge (later called <u>electrons</u>). This meant the atom was not indivisible as Dalton proposed. So, Thomson developed a new atomic model.

Taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons (with negative charge) were immersed in the atom of positive charge that counteracted the negative charge of the electrons. <u>Just like the raisins embedded in a pudding or bread. </u>

That is why this model was called the <u>raisin pudding atomic model.</u>

4 0

4 years ago

Two long, straight wires are parallel and are separated by a distance of d = 0.210 m. The top wire in the sketch carries current

love history [14]

Answer:

$1.88\cdot 10^{-5} T$ , inside the plane

Explanation:

We need to calculate the magnitude and direction of the magnetic field produced by each wire first, using the formula

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the distance from the wire

For the top wire,

I = 4.00 A

r = d/2 = 0.105 m (since we are evaluating the field half-way between the two wires)

so

$B_1 = \frac{(4\pi\cdot 10^{-7})(4.00)}{2\pi(0.105)}=7.6\cdot 10^{-6}T$

And using the right-hand rule (thumb in the same direction as the current (to the right), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is inside the plane

For the bottom wire,

I = 5.90 A

r = 0.105 m

so

$B_2 = \frac{(4\pi\cdot 10^{-7})(5.90)}{2\pi(0.105)}=1.12\cdot 10^{-5}T$

And using the right-hand rule (thumb in the same direction as the current (to the left), other fingers wrapped around the thumb indicating the direction of the magnetic field lines), we find that the direction of the field lines at point P is also inside the plane

So both field add together at point P, and the magnitude of the resultant field is:

$B=B_1+B_2 = 7.6\cdot 10^{-6} T+1.12\cdot 10^{-5}T=1.88\cdot 10^{-5} T$

And the direction is inside the plane.

3 0

3 years ago

A boat is subject to hydrodynamic drag forces of F = 40v 2 [N] where v is the magnitude of velocity in [m/s]. The boat’s mass is

stepladder [879]

Answer:

It has moved a distance, S = 25.9 m

Explanation:

F = 40 v²........(1)

$F = mv\frac{dv}{dS}$ .........(2)

Equating (1) and (2)

$-mv\frac{dv}{dS} = 40 v^2\\\\v\frac{dv}{dS} = \frac{-40 v^2}{m} \\\\m = 450 kg\\v\frac{dv}{dS} = \frac{-40 v^2}{450}\\\\ \frac{dv}{v} = \frac{40}{450} dS\\$

Integrate both sides:

$v_1 = 1, v_2 = 10$

$\int\limits^{10}_1 {\frac{1}{v} } \, dv = \frac{-40}{450} \int\limits^S_0 \, dS \\\\ln\frac{1}{10} = \frac{-40}{450} (S-0)\\\\S = \frac{-40}{450} ln(0.1)\\\\S = 25.9 m$ .

6 0

3 years ago

H e l p <br> P l e a s e.

gtnhenbr [62]

Answer:

I still cant see no matter how much I zoom in!???!

Explanation:

To Small

8 0

3 years ago

Read 2 more answers

Using what you know about newton's second law explain why a car with a lager mass use more fuel than a car with a smalle mass. a

NNADVOKAT [17]

The car with bigger Mass requires more fuel because More mass the heavier it gets. you can't grab the little car fuel tank and put it in the big car it will run out to fast.

overall answer : The bigger the mass = more weight which means it needs more fuel to run

3 0

3 years ago