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Romashka-Z-Leto [24]

3 years ago

15

ASAP WILL GIVE BRAINLIEST Take a piece of paper, some nails and a magnet bar, place a magnet at the centre of the paper, and spr

inkle the nails. Observe what will happen to the nails and what other phenomena you can observe.

1 answer:

Zanzabum3 years ago

7 0

Answer: find the answer in the explanation

Explanation:

When a magnet is placed at the centre of the paper, and the nails are sprinkled on the paper, what will happen to the nails is that, the nails will form a pattern on the paper according to the magnetic field of the bar magnetic pole.

Other phenomena you can observe are:

1.) The nails will align themselves and show some lines of forces which is equivalent to the magnetic field lines

2.) The direction of the line of forces

3.) The strength of the magnetic field pole.

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A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the

andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

$F = mg$

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

$\sum \tau = 0$

Regarding the forces we have,

$3Mg-1mg=0$

Re-arrange to find M,

$M = \frac{m}{3}$

$M = \frac{50}{3}$

$M = 16.67Kg$

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0

3 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

5 0

3 years ago

Using a crowbar, a person can remove a nail by exerting little force, whereas pulling directly on the nail requires a large forc

algol [13]

Here we deal with a lever law. It states that product of force and distance from a fixed point on a lever is equal on both sides.

F₁*d₁ = F₂*d₂

By analysing this formula we can see that applying small force on a great length equals great force on a small length.
To remove nail we need to apply certain force. If we use F₁ for this required force we can see that on other side we need to apply certain force. If we have greater arm length we need smaller force. In a crowbar arm length along which we apply force is greater than length of our arm. This leads to a conclusion that we need smaller force when using crowbar. Depending on the length of a nail it is possible that we need to apply force that is greater than force required to remove nail.

7 0

3 years ago

The image above was taken by the spacecraft Messenger as it flew by the planet Mercury. The terrain shown in this image is typic

lorasvet [3.4K]

The answer is

C

Hope this helps

7 0

3 years ago