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sertanlavr [38]
2 years ago
15

9. What is the gravitational potential energy of a 61.2 kg person standing on the roof of a 10-storey building? (Each storey is

2.50 m high.)
Physics
1 answer:
kow [346]2 years ago
3 0

Answer:

15009

Explanation:

PE = mgh

PE = 61.2(9.81)(10 * 2.50)

PE = 15009.3

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Sound travels at a rate of 340 m/s in all directions through air. Matt rings a very loud bell at one location, and Steve hears i
Vika [28.1K]

Answer:

110 m/s

Explanation:

because if you subtract 450 from 340 you get 110

6 0
2 years ago
A negative charge is moved from point A to point B along an equipotential surface. Which of the following statements must be tru
elena-s [515]

Answer:

C) No work is required to move the negative charge from point A to point B.

Explanation:

An equipotential surface is defined as a surface connecting all the points at the same potential.

Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.

The work done when moving a charge is given by

W=q\Delta V

where

q is the charge

\Delta V is the potential difference between the initial and final point of motion of the charge

However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so

\Delta V=0

And so, the work done is also zero.

7 0
3 years ago
This is the smallest unit of matter that cannot be physiologically broken down.
yuradex [85]
An atom is the smallest unit of matter. It can't be broken. It consist of dense nuleus surrounded by negatively charged electrons.
4 0
3 years ago
A star produces 2x10^26 watts. how much energy does it lose every minutes
Step2247 [10]

Answer:

Energy loss per minute will be 120\times 10^{26}j

Explanation:

We have given the star produces power of 2\times 10^{26}W

We know that 1 W = 1 J/sec

So 2\times 10^{26}W=2\times 10^{26}J/sec

Given time = 1 minute = 60 sec

So the energy loss per minute =2\times 10^{26}\times 60=120\times 10^{26}j

We multiply with 60 we have to calculate energy loss per minute

7 0
3 years ago
How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0
3 years ago
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