Which of the following is a conductor?<br><br> copper<br> water<br> aluminum<br> all of the above

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

denis23 [38]

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

3 0

2 years ago

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In order to slide a heavy desk across the floor at constant speed in a straight line, you have to exert a horizontal force of 40

san4es73 [151]

Answer:

F = f from Newton’s first law.

Explanation:

since the desk is moved in a straight line at a constant speed, newton first law tell us that the two forces must be equal.

Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. since the table has been set in motion by the 400 N force, it will remain in motion unless it is been acted upon by an external force, and this means that the 400 N must be equal to the frictional force for it to have been in motion in the first instance.

3 0

2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

2 years ago

A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the pers

jek_recluse [69]

Answer:

0.74 m/s

Explanation:

From the question,

We apply the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

Since the skateboard, the person and the brick where stationary, therefore, the total momentum before collision is 0

0 = Total momentum after collision

(m+M)V + m'v = 0

Where m = mass of the skateboard, M = mass of the person, m' = mass of the brick, V = recoil velocity of the person and the skateboard, v = velocity of the brick

make V the subject of the equation above

V = -m'v/(m+M)................... Equation 1

Given: m = 4.10 kg, M = 68.0 kg, m' = 2.50 kg, v = 21.0 m/s.

Substitute these values into equation 1

V = -(2.5×21)/(68+2.5)

V = 52.50/70.5

V = 0.74 m/s

4 0

2 years ago

A car moving at a speed of 36 km/h reaches the foot of a smooth

boyakko [2]

Answer:

d = 10.2 m

Explanation:

When the car travels up the inclined plane, its kinetic energy will be used to do the work in climbing up. So according to the law of conservation of energy, we can write that:

$Kinetic\ Energy\ of\ the \ Car = Work\ Done\ while\ moving\ up\ the\ plane\\\frac{1}{2}mv^{2} = Fd$

where,

m = mass of car

v = speed of car at the start of plane = (36 km/h)(1000 m/1 km)(1 h/3600 s)

v = 10 m/s

F = force on the car in direction of inclination = W Sin θ

W = weight of car = mg

θ = Angle of inclinition = 30°

d = distance covered up the ramp = ?

Therefore,

$\frac{1}{2}mv^{2} = mgdSin\theta\\\frac{1}{2}v^{2} = gdSin\theta\\\frac{1}{2}(10\ m/s)^{2} = d(9.81\ m/s^{2}) Sin\ 30^{0}$

4 0

2 years ago

Which of the following is a conductor? copper water aluminum all of the above