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3 years ago

Describe each of Newton’s Laws of Motion in ice skating. What can you design/develop to improve ice skating?

Physics

2 answers:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

<h3>The general low level of friction on ice allows a skater to glide along the surface smoothly without friction stopping the motion as soon as it's begun.</h3>

denis23 [38]3 years ago

3 0

Newton's three laws of motion can be used to describe the motion of the ice skating.

<h3>Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.

<h3>Newton's second law of motion</h3>

Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.

Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.

<h3>Newton's third law of motion</h3>

This law states that action and reaction are equal and opposite.

Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.

Learn more about Newton's law here: brainly.com/question/3999427

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How much heat is required to convert 500g of liquid water at 28°C into steam at 150°C? Take the specific heat capacity of water

JulijaS [17]

Answer:

1,327,063Joules

Explanation:

Heat energy is the energy needed to convert the state of a body from one phase to another.

According to the question, we want to calculate the total heat required to convert water into vapour (steam).

Note that before water can vapourize, it has to reach the boiling point first which is at 100°C. Heat energy needed to convert the water to 100°C is expressed as H1 = mcΔθ

m is the mass of the object in kg =0.5kg

c is the spcific heat capacity of water = 4183J/kg°C

Δθ is the change in temperature = 100-28 = 72°C

H1 = 0.5*4183*72

H1 = 150,588Joules

Energy required to convert the water to team H2 =mLsteam

Lsteam is the latent heat of vaporization = 2.26×10⁶J/kg

H2 = 0.5*2.26×10⁶

H2 = 1130000Joules

Heat energy needed to convert the water to 150°C is expressed as H3 = mcΔθ

m is the mass of the object in kg =0.5kg

c is the spcific heat capacity of steam= 1859J/kg°C

Δθ is the change in temperature = 150-100 = 50°C

H3 = 0.5*1859*50

H1 = 46,475Joules

Total Heat requires = H1+H2+H3 = 150,588Joules+1130000Joules+ 46,475Joules = 1,327,063Joules

8 0

3 years ago

A helicopter (m = 3250 kg) is cruising at a speed of 56.9 m/s atan altitude of 185 m. What is the total mechanical energy of the

taurus [48]

Answer:

The mechanical energy of the helicopter is $1.12\times 10^7\ J$ .

Explanation:

It is given that,

Mass of the helicopter, m = 3250 kg

Speed of the helicopter, v = 56.9 m/s

Position of the helicopter, h = 185 m

The energy possessed by an object due to its motion is called its kinetic energy. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3250\times (56.9)^2$

$E=5.26\times 10^6\ J$

The energy possessed by an object due to its position is called its potential energy. It is given by :

$E=mgh$

$E=3250\times 9.8\times 185$

$E=5.89\times 10^6\ J$

The sum of kinetic and potential energy is called mechanical energy of the system. It is given by :

$M=5.26\times 10^6+5.89\times 10^6$

$M=11.15\times 10^6\ J$

$M=1.12\times 10^7\ J$

So, the mechanical energy of the helicopter is $1.12\times 10^7\ J$ . Hence, this is the required solution.

6 0

3 years ago

How does the mass of a bowling ball that has been rolled down the lane affect the kinetic energy?

solniwko [45]

The heavier it is the more frictions it creates sliding on the floor. this make more kinetic energy.

3 0

3 years ago

Read 2 more answers

muscular movement involving the walls of the digestive tract that serve to mix materials and move them along the track

Elden [556K]

Contractions I think

5 0

3 years ago

A +26.3 uC charge qy is repelled by a force

Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0

3 years ago