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-Dominant- [34]
3 years ago
8

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

f heat could be transferred to sore muscles?
Physics
2 answers:
IgorC [24]3 years ago
8 0

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

alexandr402 [8]3 years ago
4 0
<h2>Answer:</h2>

125.007 kJ

<h2>Explanation:</h2>

The quantity of heat (Q) transferable in a heating process is the product of the mass (m) of the object involved, the specific heat capacity of the object and the change in temperature (ΔT). i.e

Q = m x c x Δ T     ------------------------(i)

Where, from the question;

m = mass of water = 787g

c = specific heat capacity of water = 4.18 J/g°C  ---- (a known constant)

ΔT = 75°C - 37°C = 38°C

<em>Substitute these values into equation (i)</em>

Q = 787 x 4.18 x 38

Q = 125007.08 J

<em>Convert to kilojoules;</em>

Q = 125.007 kJ

Therefore, the quantity of heat (in kilojoules) that could be transferred to sore muscles is 125.007

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Answer:

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The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
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Answer:

one-third of its weight on Earth's surface

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Weight of an object is = W = m*g

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Weight of probe on Mars = w₂ = m * g₁ /3

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4 years ago
a person brings his face close to a mirror . h e finds that the image of his face is magnified write the name of the mirror used
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Answer:

Spherical concave mirrors

Explanation:

Like spherical convex mirrors, spherical concave mirrors have a focus. If the object is closer to the mirror than the focal point is, the image will be virtual, like we talked about before for the plane mirror and the convex mirror.

Concave mirrors, on the other hand, can have real images. If the object is further away from the mirror than the focal point, the image will be upside-down and real---meaning that the image appears on the same side of the mirror as the object.

The closer the object comes to the focal point (without passing it), the bigger the image will be.

You can try this yourself by looking into the concave side of a shiny spoon. If you look into the spoon while holding it at arm’s length, you’ll see an extremely magnified, upside-down image of your face. But as you bring the spoon closer to your eyes, the image will get bigger and bigger.

<em>- Hope this helps! <3</em>

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3 years ago
A model rocket fired vertically from the ground ascends with a constant vertical acceleration of 52.7 m/s2 for 1.41 s. its fuel
olya-2409 [2.1K]
For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is

a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.

-52.7 = (0-v1)/1.41
v1 = 74.31 m/s

We can use v1 for the formula of the maximum height attained by an object thrown upwards:

Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m

The maximum height attained by the model rocket is 281.42 m.

For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.

Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.

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Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s

Therefore, the total time= 1.41+6.83+7.57 = 15.81 s

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How many (whole number of) 77 kg people
BabaBlast [244]
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