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1 year ago

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Which statement is true of a piece of ice at 0°C that is put into a freezer at -18°C?

1 answer:

butalik [34]1 year ago

3 0

The statement which is true about a piece of ice at 0°C which is put into a freezer at -18°C is it having the temperature of the freezer.

<h3>What is Temperature?</h3>

This is referred to the degree of hotness or coldness of a body and the unit is Celsius or Kelvin.

The ice at 0°C will experience a change in temperature of the freezer when put in it in this scenario.

Read more about Ice here brainly.com/question/2267329

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Explain how it is that you could have 2 parents with brown eyes, but you have blue eyes. Talk about the role chromosomes, genoty

steposvetlana [31]

Answer:

If both parents had a recessive allele for blue eyes and they had a brown phenotype it is completely possible for the child to have blue eyes.

Explanation:

Chromosomes play the role of transfering the encoded strands of DNA to the child. Genotypes are the traits or groups of traits transferred and need to carry the trait of blue eyes in order for the child to have them. As long as the parents had a recessive allele for blue eyes they could have a blue eyed child.

(apologies for the incomplete answer, I accidentally hit send)

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3 years ago

a 7 kilogram cat is resting on top of a bookshelf that is 2 meters high. what is the cat’s gravitational potential energy relati

attashe74 [19]

PE = mgh
Mass, m = 7kg, g ≈ 10 m/s², height = 2m

= 7*10*2

= 140 Joules.

5 0

3 years ago

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A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje

Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

$\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m$

6 0

3 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

The astronomer who was imprisoned by the church for announcing his scientific discoveries was Brahe Galileo Aristotle Copernicus

tekilochka [14]

Answer:

galileo

Explanation:

4 0

3 years ago

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