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Furkat [3]
3 years ago
7

What is the dimensional formula of potential difference​

Physics
2 answers:
nekit [7.7K]3 years ago
3 0

Answer:

The allowed energy states of a particle of mass m trapped in an infinite potential … 1 Solution By Separation Of Variables Two Dimensional. The formula for measuring potential difference is V=W/Q and this formula is known as Ohm's law.

Explanation:

patriot [66]3 years ago
3 0

Answer:

[M1 L2 T-3 I-1]

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An object has a mass of 20kg, what is its weight on Earth?
umka2103 [35]

Answer:

The formula is W=mg.

W=20×10=200.

5 0
3 years ago
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A constant force of 10 N is applied to an object. The object accelerates at a constant rate of 4 m/s2. What is the mass of the o
tatuchka [14]

Answer:

I think it is 2.5 kg.

Explanation:

I can say you do:

Fnet = m * a

So, you want to find mass:

m = Fnet / a

So, the answer you will receive is 2.5

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The regular use of a narcotic drug will lead to blank dependence
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8 0
4 years ago
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A spinning disk is rotating at a rate of 5 rad/s in the positive counterclock-wise direction. If the disk is subjected to an ang
Anit [1.1K]

Answer:

ωf = 13 rad/s

Explanation:

  • The angular acceleration, by definition, is just the rate of change of the angular velocity with respect to time, as follows:
  • α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)
  • Choosing t₀ = 0, and rearranging terms, we have

       \omega_{f} = \omega_{o} + \alpha *t  (1)

       where ω₀ = 5 rad/s, t = 4 s, α = 2 rad/s2

  • Replacing these values in (1) and solving for ωf, we get:

        \omega_{f} = 5 rad/s + (2 rad/s2*4 s) = 13 rad/s (2)

  • The wheel's angular velocity after 4s is 13 rad/s.
3 0
3 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
4 years ago
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