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Contact [7]
3 years ago
14

What is the focal length of a lens

Physics
1 answer:
goldenfox [79]3 years ago
4 0

Answer:

The focal length of a lens is refers to the distance from the center of the lens to the principal foci.

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Could we see a galaxy that is 20 billion light-years away?.
vichka [17]

Answer:

No

Explanation:

20 billion light-years away are beyond our sight and perspective on Earth and wouldn't be observable in our universe.

7 0
2 years ago
A 2.00-kilogram object weighs 19.6 newtons on earth. if the acceleration due to gravity on mars is 3.71 meters per second2, what
Harrizon [31]
Recall that mass is the amount of matter present in a body. That means it's a property that is consistent regardless of the body's current location, gravity's pull on the body, etc. 

Let's not confuse mass with weight (which is a force computed as Weight = mass x acceleration). Mass will remain constant and that means that whether the object is on Earth or on Mars, its mass remains the same.  Thus, the object will still have 2.00 kg as mass on Mars. 

Answers: 2.00 kilograms
4 0
3 years ago
Please give me answer of this question and I will mark you as brainliest ​
Degger [83]

Answer:

Maharashtra - mashru or himroo / dhoti and lugda

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Punjab - pat / kurta and pajama

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6 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
3 years ago
8.92 x 10 ^ 6 (convert from scientific notation to a standard aka regular number)
n200080 [17]
The answer is 8,920,000

7 0
3 years ago
Read 2 more answers
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