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andreyandreev [35.5K]
3 years ago
10

3. What is the potential difference across a 1 500 12 resistor carrying a

Physics
1 answer:
gregori [183]3 years ago
6 0

The potential difference across a 1500 ohm resistor carrying a current of 0.075 A is approx. equal to 113 V.

Explanation:

An ideal electrical circuit obeys Ohm's law. As per this law, the potential difference in any electrical circuit will be directly proportional to the current flowing in that circuit. And the proportionality constant is the resistance offered by the circuit.

V = I * R

As here, the current is said to be carried as 0.075 A and the resistance is 1500 ohm, then the potential difference will be

V = 0.075 * 1500 =112.5 = 113 V

Thus, the potential difference across a 1500 ohm resistor carrying a current of 0.075 A is approx. equal to 113 V.

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What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?
Assoli18 [71]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question we have

a =  \frac{70}{7}  = 10 \\

We have the final answer as

<h3>10 m/s²</h3>

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4 0
3 years ago
Which feature of a heating curve indicates a change of state?
Andreas93 [3]
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So that part of the graph is a horizontal line. 
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7 0
3 years ago
Water flows through a 2.5cm diameter pipe at a rate of
My name is Ann [436]

Answer:

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 -------------------------- eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

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d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

6 0
3 years ago
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