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Galina-37 [17]
3 years ago
13

A conversion factor is a ratio with a value of

Physics
1 answer:
Nikitich [7]3 years ago
7 0
The answer would be
B
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A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0
n200080 [17]

Answer:

a

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

b

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

    Work done   = m*g*h

where m is mass

          g is acceleration due to gravity

          and h is the height

c

Work done in terms of force and distance is = mg

where  m is mass of bucket and

            g is acceleration due to gravity  

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = \Delta P.E

W = m*g*h

where g = 9.9 m/s^2

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0
3 years ago
Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface
tatyana61 [14]

Answer:

Option 3: -48 cm

Explanation:

We are given:

refractive index; n = 1.5

radius of curvature; r2 = 24 cm

Formula for the focal length is given as;

1/f = (n - 1) × [(1/r1) - (1/r2)]

As r1 tends to infinity, 1/r1 = 0

Thus,we now have;

1/f = (n - 1) × (-1/r2)

Plugging in the relevant values;

1/f = (1.5 - 1) × (-1/24)

1/f = -0.02083333333

f = -1/0.02083333333

f = -48 cm

3 0
2 years ago
An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.
nignag [31]

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

\epsilon = -L\frac{dI}{dt}

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V

Therefore, the self-induced emf in this inductor is 4.68 mV.  

I hope it helps you!

6 0
3 years ago
How do you find the angle (to the nearest tenth) on your graph that produced the greatest range, based on the line of best fit?
VikaD [51]

Answer:

yuououiy

Explanation:

3 0
2 years ago
*science question pls answer quickly*
Mrrafil [7]
A is the answer you can believe me
3 0
2 years ago
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