When the student the sled jumps off toward the north , the sled most likely move towards the south.
<h3>What is the Newton third law?</h3>
According to the Newton third law of motion, action and reaction are equal and opposite. This means that the direction of the reaction force must also be opposite to that of the action.
As such, when the student the sled jumps off toward the north , the sled most likely move towards the south.
Learn more about Newton third law:brainly.com/question/974124
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¿El salario es un costo fijo o variable?
Los salarios anuales son costos fijos, pero otros tipos de compensación, como comisiones o horas extraordinarias, son costos variables.
The speed of the sound wave in the medium, given the data is 3900 m
<h3>Velocity of a wave </h3>
The velocity of a wave is related to its frequency and wavelength according to the following equation:
Velocity (v) = wavelength (λ) × frequency (f)
v = λf
With the above formula, we can obtain the speed of the sound wave. Details below:
<h3>How to determine speed of the sound wave</h3>
The speed of the wave can be obtained as illustrated below:
- Frequency (f) = 600 Hz
- Wavelength (λ) = 6.5 m
- Velocity (v) =?
v = λf
v = 6.5 × 600
v = 3900 m
Thus, the speed of the sound wave in the medium is 3900 m
Learn more about wave:
brainly.com/question/14630790
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Answer:
The distance between the ships is 87.84 km.
Explanation:
Given that,
Angle of first ship= 40°
Speed of first ship = 18 knots
Angle of second ship= 130°
Speed of second ship = 26 knots
We need to calculate the resultant velocity
Using cosine rule

Put the value into the formula




We need to calculate the distance between the ships

Put the value into the formula


Hence, The distance between the ships is 87.84 km.
Answer:
Explanation:
Given





R for Helium 

mass of gas 


Similarly
can be found


Work done 


Since it is a polytropic Process
therefore 






From Energy balance
Neglecting kinetic and Potential Energy change

Change in Internal Energy 




i.e. Heat is being removed