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3 years ago

When you do work on a object some of your energy is ________ to that object?

Physics

1 answer:

levacccp [35]3 years ago

5 0

<span>When you do work on a object some of your energy is "Transfers" to that object.

Hope this helps!</span>

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A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect

navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.

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3 years ago

An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the

alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

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3 years ago

Strength of electromagnets

Triss [41]

Lines of Force around an Electromagnet. ... The magnetic field strength of an electromagnet is therefore determined by the ampere turns of the coil with the more turns of wire in the coil the greater will be the strength of the magnetic field.

6 0

3 years ago

acceleration is defined as the rate of change for which characteristic? a. displacement b. position c. velocity d. time

fredd [130]

The answer is velocity.

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3 years ago

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-

nikdorinn [45]

Answer:

The answer is " $60.74^{\circ}$ ".

Explanation:

Cavity and benzene should be extended in equal quantities.

$\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\$

$\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\$

$\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\$

5 0

3 years ago