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3 years ago

Which is true of magnetic field lines, but not electric field lines

Physics

2 answers:

Fed [463]3 years ago

8 0

Answer:

B) they show which way iron shavings would align themselves

Explanation:

Stels [109]3 years ago

7 0

Answer:

B) they show which way iron shavings would align themselves

Explanation:

Let's analyze each statement:

A) they are not affected by their own source --> this is true for both magnetic and electric fields. In fact, both fields are produced (and so affected) by a source (a magnet or a current in the magnetic field case, and an electric charge in the electric field case)

B) they show which way iron shavings would align themselves --> this is only true for the magnetic field. In fact, the pieces of iron will align according to the magnetic field; however, since they are electrically neutral, they are not affected at all by an electric field.

C) they re stronger near the source and get weaker farther away --> true for both magnetic and electric fields.

D) the closer the fields lines, the stronger the fields --> also true for both magnetic and electric fields.

So, the correct answer is B.

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A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

The p wave on an electrocardiogram represents __________.

Harrizon [31]

It represents the depolarization of the atria.

<u>Explanation:</u>

The P wave speaks to the depolarization of the left and right chamber and furthermore relates to atrial compression. Carefully, the atria contract a brief moment after the P wave starts. Since it is so little, atrial repolarization is typically not unmistakable on ECG.

In an ordinary ECG, there's three particular waves. The primary wave is the P wave, which speaks to the depolarization of the atria. This happens directly before the atria agreement and drive blood into the ventricles. The following wave is known as the QRS wave.

3 0

3 years ago

I know the answer is supposed to be 20.7, but how?

Colt1911 [192]

Is the number the whole thing equal to 1?

6 0

3 years ago

A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping

leva [86]

Answer:

0.02896 kg/s

Explanation:

$A_1$ = Initial displacement = 0.5 m

$A21$ = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

$x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)$

At maximum displacement

$cos(\omega t+\phi)=1$

$\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s$

The magnitude of the damping coefficient is 0.02896 kg/s

6 0

3 years ago

Why does the sun have less gravitational pull on planet far away like Pluto?

blondinia [14]

Answer:

Because the planet is far away from the sun

Explanation:

The closer the planet is to the Sun, the greater the pull of the Sun's gravity, and the faster the planet orbits.

While, over here the Pluto is very far away from the Sun so it will have very little gravitational pull and still keeps revolving around the Sun

<h2><u><em>Pls Mark Brainliest</em></u></h2>

4 0

3 years ago