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2 years ago

15

A concave mirror of radius of 30cm produces and image three times in size of the object the distance of the object from the mirr

or is?

1 answer:

Sloan [31]2 years ago

4 0

$\\ \sf\longmapsto m=\dfrac{-v}{u}$

v is image distance and u is object distance.

$\\ \sf\longmapsto 3=\dfrac{-v}{30}$

$\\ \sf\longmapsto -v=30(3)$

$\\ \sf\longmapsto v=-90cm$

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Light travels 3,00,000km/s. ls it right velocity or speed?

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Answer:

speed

Explanation:

because speed is fast

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Answer:

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Now, if we assume time to be constant at 1 s. Then the distance covered by each car will be directly proportional to the speed of the car. Hence, the car with the greatest speed will travel the greatest distance in 1 second.

We, have:

Speed of white car > Speed of red car > Speed of green car

<u>Therefore, the white car will cover the most distance every second.</u>

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2 years ago

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2 years ago

A 60-turn coil has a diameter of 13 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.60 T so that the

kramer

Answer:

a) 0.5985 V

b) 0.05985 V

c) 0.00684 V

Explanation:

Given that

Number of turn in the coil, N = 60 turns

Magnetic field of the coil, B = 0.6 T

Diameter of the coil, d = 0.13 m

If area is given as, πd²/4, then

A = π * 0.13² * 1/4

A = 0.0133 m²

The induced emf, ε = -N(dΦ*m) /dt

Note, Φm = BA

Substituting for Φ, we have

ε = -NBA/t.

Now, we substitute for numbers in the equation

ε = -(60 * 0.6 * 0.0133)/0.8

ε = 0.4788/0.8

ε = 0.5985 V

at 8s,

ε = -(60 * 0.6 * 0.0132)/8

ε = 0.4788/8

ε = 0.05985 V

at 70s

ε = -(60 * 0.6 * 0.0132)/70

ε = 0.4788/70

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3 years ago

c) An aluminum tea kettle with mass 1.10 kg containing 1.80 kg of water is placed on a stove. If no heat is lost to the surround

True [87]

Answer:

Q = 640425 J = 640.425 KJ

Explanation:

From the law of conservation of energy:

Heat Addition = Heat Gain of Water + Heat Gain of Aluminum Kettle

$Q = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

Q = Heat addition required = ?

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$m_a$ = mass of aluminum kettle = 1.1 kg

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$C_a$ = specific heat capacity of aluminum kettle = = 890 J/kg.°C

$\Delta T_w$ = $\Delta T_a$ = change in temperature of water and kettle = 95°C - 20°C = 75°C

Therefore,

$Q = (1.8\ kg)(4200\ J/kg.^oC)(75^oC)+(1.1\ kg)(890\ J/kg.^oC)(75^oC)\\$

Q = 567000 J + 73425 J

<u>Q = 640425 J = 640.425 KJ</u>

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3 years ago