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3 years ago

To create electromagnetic induction, which of the following has to move?

Physics

2 answers:

Alexus [3.1K]3 years ago

4 0

Answer:

the answer is definitely D

kenny6666 [7]3 years ago

4 0

So the following answer are A. a wire B. A magnet C. A transformer D. either a or b

The answer is D

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The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0

2 years ago

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = $\frac{1}{4}$ F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.

6 0

3 years ago

How are base units and derived units related?

Vlad1618 [11]

Answer:

SI derived units

Other quantities, called derived quantities, are defined in terms of the seven base quantities via a system of quantity equations. The SI derived units for these derived quantities are obtained from these equations and the seven SI base units.

Explanation:

Hope this helps :D

8 0

3 years ago

Interactive Solution 11.13 presents a model for solving this problem. A solid concrete block weighs 100 N and is resting on the

mash [69]

Answer:

The value is $} N = 66 \ blocks$

Explanation:

From the question we are told that

The weight of the block is $W_b = 100 \ N$

The dimension of the block is $d = 0.400 m \ X \ 0.250 \ m \ X \ 0.130 \ m$

Generally two atmosphere is equivalent to

$P_{2atm} = 2 * 1.013 *10^{5} = 202600 \ N/m^2$

Generally 1 atm = $1.013 *10^{5} N/m^2$

The area of the block would be evaluated using width and height because we need for the smaller surface to be in contact with the ground in order to maximize the pressure and minimize number of blocks

$A = 0.250 * 0.130$

=> $A = 0.0325 \ m^2$

Generally the force due to this blocks is mathematically represented as

$F = N * W_b$

Here N is the number of blocks

$} 202600 = \frac{N * 100 }{ 0.0325}$

=> $} N = 66 \ blocks$

3 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago