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Ierofanga [76]
3 years ago
15

A fire truck has a searchlight with a resistance of 60 (ohm) which is placed across a 24-V battery. What is the current in this

circuit
Physics
1 answer:
Mariulka [41]3 years ago
3 0

Answer: 0.4 Amps

Explanation:

Voltage of battery = 24 Volts

Current I = ?

Resistance of searchlight (R)= 60ohms (Ω is the symbol for ohms)

Then, apply the formula for ohms law

Voltage = Current x resistance

i.e V = I x R

24V = I x 60Ω

I = 24V / 60Ω

I = 0.4 Amps (Amps is the unit of current)

Thus, the current in the circuit is 0.4 Amps

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Answer:

71%

Explanation:

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According to the Heisenberg uncertainty principle, if the uncertainty in the speed of an electron is 3.5 × 103 m/s, the uncertai
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Explanation:

It is given that,

Uncertainty in the speed of an electron, \Delta v=3.5\times 10^3\ m/s

According to Heisenberg uncertainty principle,

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\Delta x is the uncertainty in the position of an electron

Since, \Delta p=m\Delta v

\Delta x=\dfrac{h}{4\pi.m \Delta v}

\Delta x=\dfrac{6.6\times 10^{-34}}{4\pi\times 9.1\times 10^{-31}\times 3.5\times 10^3}

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2 years ago
A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction
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The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

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Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

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Have a nice day!

6 0
3 years ago
A 30 kg child went down a 10 m tall slide. Assuming no energy was lost as friction, what was the child's velocity when he reache
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3 years ago
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener
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Answer:

Explanation:

From the given information:

The initial PE (PE)_i = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = P.E_f -P.E_i

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ΔP.E = -P.E_i

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this can be re-written as:

\Delta U =- (-\Delta P.E_i) - \Delta K.E

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

\Delta U =(70\%) \Delta P.E_i-0

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Thus, the magnitude of the increase is = 343.35 J

7 0
3 years ago
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