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2 years ago

The black hole is___ times smaller that the star. I need answers please

Physics

1 answer:

mojhsa [17]2 years ago

3 0

Answer:

The answer is 24 (for the first question).

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>

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I need the answer to number #23 please will give BRAINLIEST

Murrr4er [49]

Yeah lowkey i think it’s B

3 0

3 years ago

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$

Where $c_p$ represent the specific heat for the water and this value from tables we know that $c_p =1 \frac{cal}{gr C}$ for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

8 0

3 years ago

Read 2 more answers

A glass rod and a steel rod are of equal length at 0C. At 100C they differ in length by

NeX [460]

The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

Learn more about linear expansion here:

brainly.com/question/14089545

6 0

2 years ago

If you dribble a basketball with a frequency of 1.77 Hz, how long does it take for you to complete 12 dribbles?

Licemer1 [7]

<h2>It takes 6.78 seconds to complete 12 dribbles.</h2>

Explanation:

Frequency of dribble = 1.77 Hz

That is

Number of dribbles in 1 second = 1.77

$\texttt{Time taken for 1 dribble = }\frac{1}{1.77}=0.565s$

Now we need to find how long does it take for you to complete 12 dribbles.

Time taken for 12 dribbles = 12 x Time taken for 1 dribble

Time taken for 12 dribbles = 12 x 0.565

Time taken for 12 dribbles = 6.78 seconds

It takes 6.78 seconds to complete 12 dribbles.

8 0

3 years ago

A perpetual-motion machine can never be built because it is not possible to eliminate...

bazaltina [42]

The answer is: [C]: "elasticity" .
________________________________________

4 0

3 years ago

Read 2 more answers