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Stella [2.4K]
3 years ago
5

You have a small piece of iron at 25 °C and place it into a large container of water at 75 °C. Which of these could be the tempe

rature of the water after 10 minutes
Physics
1 answer:
Fofino [41]3 years ago
8 0

hello^^ your answer would be 30 degrees celcuis, have a good day!

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What is the most common kind of element in the solar wind?
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8 0
3 years ago
A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th
enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

  • <em>Mass of the child, m = 16 kg</em>
  • <em>Length of the incline, L = 2 m</em>
  • <em>Angle of inclination, θ = 20⁰</em>

The vertical height of fall of the child from the top of the incline is calculated as;

sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m

The gravitational potential energy of the child at the top of the incline is calculated as;

P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0
2 years ago
4. Ano ang matinding suliranin ang kakaharapin kapag patuloy ang pagkasira ng
jasenka [17]

Answer:

I think its D

Explanation:

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5 0
3 years ago
Read 2 more answers
Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

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3 years ago
What does it mean to do science
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