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3 years ago

9

A wire loop is placed inside a magnetic field. the magnetic flux is maximized when the angel between the field lines and the sur

face of the area of the wire loop is

1 answer:

klasskru [66]3 years ago

6 0

When magnetic field lines and surface area are perpendicular to each other.

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How long will it take by 50W heater to melt 100g of ice at 0degreeC? Specific heat capacity of water = 4.2J/ (g0C), latent heat

vaieri [72.5K]

Answer:

420 s

Explanation:

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3 years ago

Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l

OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

<u>F = 85696.5 N = 85.69 KN</u>

8 0

3 years ago

A carnot heat engine receives 600 kj of heat from a source of unknown temperature and rejects 175 kj of it to a sink at 20°c. de

oee [108]

(b) 71%

The thermal efficiency of a Carnot heat engine is given by:

$\eta = \frac{W}{Q_{in}}$

where

W is the useful work done by the engine

$Q_{in}$ is the heat in input to the machine

In this problem, we have:

$Q_{in}=600 kJ$ is the heat absorbed

$W=600 kJ-175 kJ=425 kJ$ is the work done (175 kJ is the heat released to the sink, therefore the work done is equal to the difference between the heat in input and the heat released)

So, the efficiency is

$\eta = \frac{425 kJ}{600 kJ}=0.71 = 71\%$

(a) $737^{\circ}C$

The efficiency of an engine can also be rewritten as

$\eta = 1-\frac{T_C}{T_H}$

where

$T_C$ is the absolute temperature of the cold sink

$T_H$ is the temperature of the source

In this problem, the temperature of the sink is

$T_C = 20^{\circ}C + 273=293 K$

So we can re-arrange the equation to find the temperature of the source:

$T_H = \frac{T_C}{1-\eta}=\frac{293 K}{1-0.71}=1010 K\\T_H = 1010 K - 273=737^{\circ}C$

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3 years ago

Who was the first person to walk on the moon ?

liq [111]

Answer:

god correct correct

Explanation:

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4 years ago

Read 2 more answers

A circular loop of wire lies flat on a level table top. A bar magnet is held stationary above the circular loop with its north p

chubhunter [2.5K]

The direction of the induced current cannot be determined from the given information.

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4 years ago