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Vesnalui [34]
3 years ago
12

Describe a method for cooking an egg that uses conduction, convection, and radiation.

Physics
1 answer:
mestny [16]3 years ago
3 0
Steaming the egg

The heat is conduction through the pan and into the water
There is convection current in the water and the pot
The heat is radiated from the water to the egg
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an electric motor converts electrical energy into which kind of energy? potential kinetic storage chemical​
liubo4ka [24]

Answer:

kinetic

Explanation:

6 0
3 years ago
Read 2 more answers
It takes 120 minutes for a man to ride his bicycle up the road to Alpe d'Huez in France. The vertical height of the climb is 1,1
Greeley [361]
First solve the potential energy of the biker. using the fomula:
PE = mgh
where m  is the mass of the object
g is the acceleration due to gravity ( 9.81 m/s2)
h is the height

PE = 96 kg ( 1120 m ) ( 9.81 m/s2)
PE = 1054771.2 J
then power = Work / time
P = 1054771.2 J / ( 120 min ) ( 60 s / 1 min)
P = 146.5 W

8 0
4 years ago
At 9:13AM a car is traveling at 35 miles per hour. Two minutes later, the car istraveling at 85 miles per hour. Prove that are s
andreyandreev [35.5K]

Answer:

There is at least one instant which instantaneous acceleration is equal to average acceleration. a = 1500\,\frac{km}{h^{2}}.

Explanation:

The average acceleration experimented by the car is:

\bar a = \frac{85\,\frac{km}{h} - 35\,\frac{km}{h} }{\frac{2}{60}\,h }

\bar a = 1500\,\frac{km}{h^{2}}

According to the Rolle's Theorem, there is at least one instant t so that instantaneous acceleration equal to average acceleration for the analyzed interval. That is to say:

v'(c) = \frac{v(\frac{2}{60} )-v(0)}{\frac{2}{60}-0}

If car is accelerating at constant rate, instantaneous acceleration coincides with average acceleration for all instant t. Then, instantaneous acceleration is:

a = 1500\,\frac{km}{h^{2}}

6 0
3 years ago
Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
3 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity

\lambda = Wavelength

Our value are given by,

v = 345m/s

\lambda = 63cm

Replacing

f = \frac{345}{0.63}

f = 547.61Hz

Therefore the frequency of the tuning fork is 547.61Hz

6 0
3 years ago
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