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Oliga [24]
2 years ago
6

You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.

And weighing Polly, you find her mass to be 60 kg. Now, subtracting the heat warming the water from the surrounding environment, we assume Polly gave up only 5000 kJ of thermal energy to warm the bath. With these numbers, determine Polly's specific heat in units J/(kgoC).
Physics
1 answer:
ankoles [38]2 years ago
5 0

Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

Mass = 60 Kg

Energy = 5000 J

We need to calculate the specific heat

Using formula of energy

Q= mc\Delta T

c =\dfrac{Q}{m\Delta T}

Put the value into the formula

c=\dfrac{5000}{60\times(37-13)}

c=3.47222\ J/kg^{\circ}C

Hence, The specific heat is 3.47222 J/kg°C.

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An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
2 years ago
Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving
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Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

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Honeybees can see light in the ________ range of the electromagnetic spectrum.
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Humans can see wavelengths in the visible part of the electromagnetic spectrum. That is the range of approximately 400 - 700 nm. Honeybees can see visible light and about 100 nm more in the ultraviolet part of the electromagnetic spectrum. That is approximately 300 - 700 nm. 
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A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri
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at = r * aa  
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In this vLab you used a complex machine to launch a projectile with the ultimate goal of hitting the target. Assume you built a
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For a catapult to fire a projectile a significant range, the projectile will need a large mass. The machine would have to be very large to compensate for that. Also, the machine would be highly inaccurate. It would be entirely too difficult to pinpoint the exact location in which the projectile will hit. If you were to use a projectile that had a smaller mass, it would too easily be affected by friction, wind, and other outside forces. The machine used to fire the projectile itself, would have to be large, and it would be very inefficient.
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