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dybincka [34]
3 years ago
7

The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 2.67 from 8.19 Wb in a time of 0.0386 s. The

average induced current in the coil is 275 A. What is the resistance of the wire?
Physics
1 answer:
Alik [6]3 years ago
3 0
I believe that the resistance of the wire is 12
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I need help please somebody help me
Morgarella [4.7K]

Answer:

Generally, magnets are attracted to objects that are made of the metals iron, nickel, or cobalt. These materials are called ferromagnetic materials. ... When all or most of the domains are aligned in the same direction, the whole object becomes magnetized in that direction and becomes a magnet.

Explanation:

5 0
2 years ago
Why are there multiple versions of the scientific method?
gavmur [86]

Answer:

it's because some versions have more steps and others have less

4 0
2 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
_____ are pictures of relationships
butalik [34]

The blank in the question can be filled with the word, “Graph”. Therefore, Graphs are the pictures which are in relationships.

<u>Explanation: </u>

Graph usually represents a set of data which is nonlinear in occurrence and has some relationship between the two given data. And as graph are pictorial representation, it is simply assumed as the pictures of relationships.

For example, a graph can be drawn for the set of data for the presence of number of students of all the sections of the particular class of a school, as they are relative. But making the graph for number of students in all section of all class but different school cannot be done as non-relative.

4 0
3 years ago
Read 2 more answers
A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a
Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

F=ma      Eqn. (1)

where

F : is the Net Force in Newtons ( N )

m : is the mass ( kg )

a  : is the acceleration ( m/s^2 )

We also know that the acceleration is denoted by the velocity ( v ) of an object as a function of time ( t ) with

a=\frac{v}{t}         Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

F=m\frac{v}{t}\\ \\Ft=mv     Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. Ft is in fact the Impulse  J of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ 22N and t=4 sec (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

J=Ft\\J=(22N)(4sec)\\J=88 kg m/s

So the impulse of the cart is J=88kg m/s

Then, we know that the cart is moving at v=10 m/s. Plugging in the values in Eqn. (3) we have:

(22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg

So the mass of the cart is m=8.8kg.

8 0
3 years ago
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