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2 years ago

14

How is an electromagnet different from a bar magnet?

2 answers:

liberstina [14]2 years ago

7 0

C. is the answer please mark me brainliest I really need it

Reptile [31]2 years ago

5 0

The answer for the question is c

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Figure 2. sketch of bar and magnetic field lines observations PLEASE HELPPPP

labwork [276]

Hope this sketch helps

5 0

2 years ago

A person travelled 350 m east from his home and returns back home an hour has displacement of_?

Svetradugi [14.3K]

Answer:

vector of zero magnitude

Explanation:

The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.

In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.

If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude

7 0

2 years ago

It takes a ball 30 seconds to travel from point A to point B. What is the ball’s speed?

AlladinOne [14]

Direction from A to B divided by time(30s)

3 0

3 years ago

Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

brilliants [131]

Answer:

Speed will be 30810 rpm

Explanation:

We have given diameter of the tire d = 24 inch

So radius $r=\frac{d}{2}=\frac{24}{2}=12imch$

We have given linear velocity v = 35 mph

We know that linear velocity is given by $v=\omega r$

$35=\omega \times 12$

$\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min$

As we know that 1 mile = 63360 inch and 1 hour = 60 min

3 0

3 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0

3 years ago