2 years ago

In what circuit does an increase in resistance mean an increase in current?

Physics

1 answer:

skad [1K]2 years ago

6 0

I think in parallel circuits.

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A 1000 kg train car traveling at 12 m/s on a level track has 400 kg of sand dumped, vertically, into it. What is the final speed

crimeas [40]

Answer:

The final speed of the train car is 8.57 m/s.

Explanation:

Given that,

Mass of train car = 1000 kg

Mass of sand = 400 kg

Speed of train car = 12 m/s

We need to calculate the final speed of the train car after the sand is dumped

Using conservation of momentum

$m_{1}u_{1}=(m_{1}+m_{2})v$ ...(I)

Put the value in the equation (I)

$1000\times12=(1000+400)v$

$v=\dfrac{1000\times12}{1400}$

$v=8.57\ m/s$

Hence, The final speed of the train car is 8.57 m/s.

5 0

3 years ago

Which is a chemical property that can be used to identify calcium carbonate?

Zina [86]

I think the answer is d i think

5 0

3 years ago

When a 12.0-V battery causes 2.00 μC of charge to flow onto the plates of an air-filled capacitor, how much work did the battery

Mrac [35]

Answer:

Explanation:

Voltage, V = 12 V

Charge, q = 2 micro coulomb = 2 x 10^-6 C

Work = energy

W = 0.5 x q x V

W = 0.5 x 2 x 10^-6 x 12

W = 12 x 10^-6 J

7 0

4 years ago

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A single proton has which electrical charge?

valentina_108 [34]

The answer is B- a positive charge

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3 years ago

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1. A 2,000-turn solenoid is 65 cm long and has cross-sectional area 30 cm2. What rate of change of current will produce a 600 Vo

JulsSmile [24]

Answer:

$\frac{dI}{dt} = 2.59\ x\ 10^4\ A/s$

Explanation:

First, we will calculate the inductance of the solenoid by using the following formula:

$L = \frac{\mu_o AN^2}{l}$

where,

L = self-inductance of solenoid = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

A = Cross-sectional area = 30 cm² = 3 x 10⁻³ m²

N = No. of turns = 2000

l = length = 65 cm = 0.65 m

Therefore,

$L = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(3\ x\ 10^{-3}\ m^2)(2000)^2}{0.65\ m}\\\\L = 0.0232\ H$

Now, we will use Faraday's law to calculate the rate of change of current:

$emf = L\frac{dI}{dt}\\\\ \frac{dI}{dt} =\frac{emf}{L} \\\\ \frac{dI}{dt} =\frac{600\ V}{0.0232\ H}\\\\ \frac{dI}{dt} = 2.59\ x\ 10^4\ A/s$

5 0

3 years ago