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2 years ago

2. A 1.54 kΩ resistor is connected to an AC voltage source with an rms voltage of 240 V.

Physics

1 answer:

svp [43]2 years ago

8 0

(a) The maximum potential difference across the resistor is 339.41 V.

(b) The maximum current through the resistor is 0.23 A.

(d) The average power dissipated by the resistor is 38.4 W.

<h3>Maximum potential difference</h3>

Vrms = 0.7071V₀

where;

V₀ is peak voltage

V₀ = Vrms/0.7071

V₀ = 240/0.7071

V₀ = 339.41 V

<h3> rms current through the resistor </h3>

I(rms) = V(rms)/R

I(rms) = (240)/(1,540)

I(rms) = 0.16 A

<h3>maximum current through the resistor </h3>

I₀ = I(rms)/0.7071

I₀ = (0.16)/0.7071

I₀ = 0.23 A

<h3> Average power dissipated by the resistor</h3>

P = I(rms) x V(rms)

P = 0.16 x 240

P = 38.4 W

Learn more about maximum current here: brainly.com/question/14562756

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Lab: newton's laws of motion assignment: lab report

poizon [28]

Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

<h3>What are Newton's motion laws?</h3>

Newton's motion laws are a set of scientific statements aimed at explaining the physical property of movement.

These laws explain why objects in movement tend to maintain the same velocity for a short period of time.

In conclusion, Newton's motion laws state that if an object is at rest or in movement, it will tend to maintain its basal state.

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1 year ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

$\dfrac{mg}{2} = \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

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2 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

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3 years ago

Gabriella got a burn on her arm from a chemical spill and rinsed the area thoroughly with water. Which piece of safety equipment

ioda

I believe it she should use the first aid kit next

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3 years ago

Read 2 more answers

How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

stealth61 [152]

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

$W = P\Delta V$

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

$W = P(0\ m^3)\\$

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3 years ago