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3 years ago

Hyperthermic body temperature can range from 41.0°C to 42.3°C. What is this range in degrees Fahrenheit?

Physics

2 answers:

Gemiola [76]3 years ago

4 0

Answer:

To convert celcius to farenheit the formula to be used is (c*9/5)+32

we have to use this formula and we will get the final answer then do the same with the other no. also then subtract both the answer to get the final answer

IRISSAK [1]3 years ago

3 0

Answer:

Hyperthermic body:

A body which is unable to dissipate the energy that it posses or contains inside itself. As there are various techniques through which any normal body can remove the extra amount of heat from itself and maintain a balance body temperature,T.

As, we have the following formula to convert the temperature,T present in centigrade to the Fahrenheit, as follows;

T(Fahrenheit)=T(centigrade)×1.8+32,

Explanation:

So by putting the values provided in the problem, we can now resolve it as,

For the lower limit we can have,

T(Fahrenheit)=T(41)×1.8+32,
T(Fahrenheit)=105.8°F.

And for the higher limits we can resolve them as,

T(Fahrenheit)=T(42.3)×1.8+32,
T(Fahrenheit)=108.14°F.

As we have the hyperthermic body temperature range from the 105.8°F to 108.14°F.

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2 years ago

Calculate the amount of work done if you use a 100N force to push a 50kg box 5m across the kitchen floor.

irinina [24]

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3 years ago

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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

Arisa [49]

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

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3 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$