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telo118 [61]
3 years ago
8

A converging lens forms an image at 43 cm. A 9 cm tall object is placed in front of the lens at a distance of 90 cm.

Physics
1 answer:
77julia77 [94]3 years ago
3 0
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A boy of mass 50 kg exerts a pressure of 25000 pa calculate the area
gtnhenbr [62]

Answer: A = 0.02 m²

Explanation: Solution:

Pressure is expressed in the formula:

P = F / A       where F = mg

  = mg /A

  = 50 kg ( 9.8 m/s²) / 25000 Pa

  =  0.02 m²

3 0
3 years ago
How do you find the rest mass (kg) of a 3.1 eV electron?
Scilla [17]

Answer:

Explanation:

The rest energy of any substance is defined by the Einstein's mass energy equivalence relation. Thus the rest mass of a electron is 9.11x10^-31 kg. The speed of light is 299,792,458 m/s. Thus multiplying the square of speed of light with the rest mass of electron gives the rest energy of the electron.

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3 years ago
A person runs up several flights of stairs and is exhausted at the top. Later the same person walks up the same stairs and does
KIM [24]

Answer:

Explanation:

Work done by the person depends on its mass and the height raised. It also depends on the acceleration due to gravity.

As the height raised and the mass of person is same. The value of acceleration due to gravity is also constant So, the work done is also same in both the cases.

6 0
3 years ago
A man pulls a 10 kg box at constant speed across the floor with a rope. The tension in the rope is 120 N at an angle of 30°.
Murrr4er [49]

Answer:

98n

Explanation:

5 0
3 years ago
Consider a spherical planet of uniform density rho. The distance from the planet's center to its surface (i.e., the planet's rad
alisha [4.7K]

Answer:

a) g(r) = 4\pi \cdot G \cdot \rho\cdot r, b) g = 4\pi \cdot G \cdot \rho\cdot r_{P}

Explanation:

a) The acceleration due to gravity inside the planet is:

dg = G\cdot \frac{\rho \cdot dV}{r^{2}}

dg = G\cdot \frac{\rho \cdot dV}{r^{2}}

dg = G\cdot \frac{4\pi\cdot \rho \cdot r^{2}\,dr}{r^{2}}

dg = 4\pi\cdot G\cdot \rho \,dr

g(r) = 4\pi \cdot G \cdot \rho\cdot r

b) The acceleration at the surface of the planet is:

g = 4\pi \cdot G \cdot \rho\cdot r_{P}

5 0
3 years ago
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