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amm1812
3 years ago
12

To turn a coil of wire into a magnet, run a(n) ____ through it.

Physics
2 answers:
evablogger [386]3 years ago
7 0

A.) Electromagnetic Current

please mark me as the brainliest

Llana [10]3 years ago
3 0
B. Magnetic field im so bad at these
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What type of change occurs when water changes from a solid to a liquid
Jobisdone [24]

What type of change occurs when water changes from solid to a liquid a phase change a physical change and irreversible change both A and the
8 0
3 years ago
A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute
levacccp [35]

Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

          m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

where,  m_{1} = mass of the projectile

            m_{2} = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

         75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v

                                  = \frac{45}{50.075}

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E = \frac{1}{2}mv^{2}

                  = \frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}

                  = 13500 J

Now, energy after the impact will be as follows.

             E' = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

           \Delta E = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n = \frac{\Delta E}{E}

             = \frac{13480}{13500} \times 100

             = 99.85

             = 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0
3 years ago
Read 2 more answers
Explain the correspondence that lets us easily translate between linear motion and rotational motion. What are the linear analog
RideAnS [48]

Explanation:

The linear analog of angle is angle itself.

The linear analog of angular velocity is linear velocity.

ω is angular velocity, therefore linear velocity is given by v

∴ for linear velocity, v^{2} = u^{2}+2.a.S

   for angular velocity, \omega_{f}^{2}  = \omega _{i}^{2}+2.a.S

The linear analog of angular acceleration is acceleration.

α is angular acceleration whereas as a is linear acceleration.

∴ for linear acceleration, v = u + a.t

  for angular acceleration, \omega_{f}= \omega _{i}+\alpha .t

The linear analog of moment of inertia is mass.

I is moment of inertia and m is mass,

∴ for linear analog, F = m.a

  for angular analog, τ - I.α

4 0
3 years ago
Jan ran 4 miles north in 28 minutes. What was Jan's average velocity?
fenix001 [56]

Answer:

3.83 m/s

Explanation:

Given that,

Distance covered by Jan, d = 4 miles

1 mile = 1609.34 m

4 miles = 6437.38 m

Time, t = 28 minutes = 1680 s

Jan's average speed,

v = d/t

v=\dfrac{6437.38\ \text{m}}{1680\ \text{s}}\\\\v=3.83\ \text{m/s}

Hence, the average velocity of Jan is 3.83 m/s.

6 0
3 years ago
What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0
3 years ago
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