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Y_Kistochka [10]
3 years ago
14

when a electric current is passed through an insulated wire that is coiled around an iron core like a nail what is created?

Physics
1 answer:
Anika [276]3 years ago
8 0

electromagnet

When a electric current is passed through an insulated wire that is coiled around an iron core, like a nail, an electromagnet is created.

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Which of these will be the correct relationship between work input and work output?
dsp73

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

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2 years ago
Mountains are part of which earth sphere?? Am I correct?
sammy [17]
You are correct. Mountains are part of the lithosphere.
8 0
3 years ago
Read 2 more answers
An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force
sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = 10ms^{-2}, F = ma = 15*10 = 150N

2) m = 3kg, v = 4ms^{-1}, r = 4m, F = \frac{mv^{2} }{r}

\frac{3*4^{2} }{4} = 12N

3) a = 10ms^{-2}, r = 10m, v=?

F = \frac{mv^{2} }{r} and F = ma

equating the two equations and cancelling a, we have:

\frac{v^{2} }{r} = a

making v the subject of formula, we have:

v = \sqrt{ar}

= \sqrt{100}

= 10ms^{-1}

4) r = 10m, v = 5ms^{-1}, a = ?

F = \frac{mv^{2} }{r}

F = ma

equating the above equations and making a subject of formula, we get:

a = \frac{v^{2} }{r}

a = 25/10 = 2.5ms^{-2}

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is tripled

F = (3v)^{2}

F = 9v^{2}

We can see that the force will be 9X greater than it was.

9) F = \frac{mv^{2} }{r}

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = v^{2}

Now, if v is doubled

F = (2v)^{2}

F = 4v^{2}

We can see that the force will be 4X greater than it was.

10) F = \frac{mv^{2} }{r}

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0
3 years ago
At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She
Citrus2011 [14]

Answer:

a

The height is   H  = 6.74 \  m

b

The horizontal distance is  D  = 23.74 \  m

Explanation:

From the question we are told that

  The speed is  v  =  15 \  m/s

  The angle is  \theta  =  40^o

   The height of the cannon from the ground is  h  =  2 m

  The distance of the net from the ground is k  =  1 m

 

Generally the maximum height she reaches is mathematically represented as  

     H  =  \frac{v^2 sin^2 \theta }{2 *  g }  +  h

=>    H  =  \frac{(15)^2 [sin (40)]^2 }{2 * 9.8}  +  2

=>    H  = 6.74 \  m

Generally from kinematic equation  

    s = ut + \frac{1}{2} at^2

Here s is the displacement which is mathematically represented as

         s  =  [-(h-k)]  

    =>  s =  -(2-1)

    =>  s  = -1 m

There reason why s =  -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

     a =  -g = -9.8

    u  =  v sin (\theta)

So

    -1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2

=>  -4.9t^2 + 9.6418t + 1 = 0

using  quadratic formula to solve the equation we have

    t  =  2.07 \  s

Generally distance covered along the horizontal is  

   D  =  v cos (40) *  2.07

=>   D  =  15 cos (40) *  2.07

=>   D  = 23.74 \  m

7 0
3 years ago
An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste
sesenic [268]

Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

According to work energy theorem work done by all the forces is equal to change in kinetic energy of object

W=\frac{1}{2}mv^2---1

where m=mass of object

v=velocity of object

When the object is already have velocity v then the final speed is given by work energy theorem

W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2

From 1 and 2 we get

\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2

2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2

v_f^2=2v^2

v_f=\sqrt{2}v                

8 0
3 years ago
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