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2 years ago

Why doesn't the skater ever regain his potential energy?

Physics

1 answer:

liq [111]2 years ago

7 0

Answer:

Friction

Explanation:

As the skateboarder goes down the ramp, potential energy is converted to kinetic energy. Because of friction, some of the energy in the system is converted to heat energy. Once the kinetic energy is converted to heat, the energy cannot be converted back to the potential or kinetic energy in the system.

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A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car

dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is option B.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

$Normal force = Gravitational force - centripetal force$

At the foot of a round hill

$Normal Force = centripetal force + Gravitational force$

4 0

4 years ago

Find the direction (clockwise or counterclockwise) of the current induced in loop AA if II is steadily decreasing. The current i

IrinaVladis [17]

Answer: counterclockwise

Explanation:

As the area of the loop decreases steadily, the flux through the loop also decreases. By Lenz’s law, any induced current will tend to oppose the decrease. Using Flemming right hand rule (Fleming's right-hand rule which shows the direction of induced current) we know that magnetic field inside the loop due to a counterclockwise current comes out of the plane. Therefore a counterclockwise current will create a stronger magnetic field inside

the loop, tending to increase the flux.

When a wire loop is moved in the direction of the current or a wire loop is being pulled through a uniform magnetic field there would be no induced current ( current loop is said to be zero)

4 0

3 years ago

Can someone please answer this, ill give you brainliest Would be very appreciated.

elixir [45]

Answer:

c because

Explanation:

Plants and trees couldn't thrive without capillary action. Capillary action helps bring water up into the roots. With the help of adhesion and cohesion, water can work it's way all the way up to the branches and leaves.

6 0

2 years ago

Read 2 more answers

This homework is meant to be in for today first lesson, please help!!!

Black_prince [1.1K]

The agriculture land needed to grow the willow is 1500000 Hectares

<h3>What is Power?</h3>

Power can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).

A coal burning power station burns 6 million tonnes of coal per kg. Coal has a average energy value of 29.25 MJ per year. Wood chip from willow trees has an energy value of 13 MJ per kg. A hectare of agricultural land can produce 9 tonnes of drt willow wood per year.

The coal consumption per year = 29.25 x 1000 x 6000000

=175500000000 MJ

If this power station burned dry willow wood instead of coal, agriculture land would be needed to grow the willow,

1 hectare of willow will produce 9 x 13x 1000 = 117000 MJ.

The agriculture land needed to grow the willow is

= 175500000000 MJ / 117000 MJ.

= 1500000 Hectares.

Thus, agriculture land needed is 1500000 Hectares.

Learn more about power.

brainly.com/question/15120631

#SPJ1

4 0

2 years ago

Determine the ratio β = v/c for each of the following.

nlexa [21]

Answer:

a) $\beta = 1.111\times 10^{-7}$ , b) $\beta = 9\times 10^{-7}$ , c) $\beta = 3.087\times 10^{-6}$ , d) $\beta = 2.5\times 10^{-5}$ , e) $\beta = 0.5$ , f) $\beta = 0.877$

Explanation:

From relativist physics we know that $c$ is the symbol for the speed of light, which equal to approximately 300000 kilometers per second. (300000000 meters per second).

a) A car traveling 120 kilometers per hour:

At first we convert the car speed into meters per second:

$v = \left(120\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 33.333\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 33.333\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{33.333\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 1.111\times 10^{-7}$

b) A commercial jet airliner traveling 270 meters per second:

The ratio $\beta$ is now calculated: ( $v = 270\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{270\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 9\times 10^{-7}$

c) A supersonic airplane traveling Mach 2.7:

At first we get the speed of the supersonic airplane from Mach's formula:

$v = Ma\cdot v_{s}$

Where:

$Ma$ - Mach number, dimensionless.

$v_{s}$ - Speed of sound in air, measured in meters per second.

If we know that $Ma = 2.7$ and $v_{s} = 343\,\frac{m}{s}$ , then the speed of the supersonic airplane is:

$v = 2.7\cdot \left(343\,\frac{m}{s} \right)$

$v = 926.1\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 926.1\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{926.1\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 3.087\times 10^{-6}$

d) The space shuttle, travelling 27000 kilometers per hour:

At first we convert the space shuttle speed into meters per second:

$v = \left(27000\,\frac{km}{h} \right)\times \left(1000\,\frac{m}{km} \right)\times \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$v = 7500\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 7500\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{7500\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 2.5\times 10^{-5}$

e) An electron traveling 30 centimeters in 2 nanoseconds:

If we assume that electron travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 0.3\,m$ and $t = 2\times 10^{-9}\,s$ , then speed of the electron is:

$v = \frac{0.3\,m}{2\times 10^{-9}\,s}$

$v = 1.50\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 1.5\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{1.5\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.5$

f) A proton traveling across a nucleus (10⁻¹⁴ meters) in 0.38 × 10⁻²² seconds:

If we assume that proton travels at constant velocity, then speed is obtained as follows:

$v = \frac{d}{t}$

Where:

$v$ - Speed, measured in meters per second.

$d$ - Travelled distance, measured in meters.

$t$ - Time, measured in seconds.

If we know that $d = 10^{-14}\,m$ and $t = 0.38\times 10^{-22}\,s$ , then speed of the electron is:

$v = \frac{10^{-14}\,m}{0.38\times 10^{-22}\,s}$

$v = 2.632\times 10^{8}\,\frac{m}{s}$

The ratio $\beta$ is now calculated: ( $v = 2.632\times 10^{8}\,\frac{m}{s}$ , $c = 3\times 10^{8}\,\frac{m}{s}$ )

$\beta = \frac{2.632\times 10^{8}\,\frac{m}{s} }{3\times 10^{8}\,\frac{m}{s} }$

$\beta = 0.877$

4 0

3 years ago

Why doesn't the skater ever regain his potential energy? ​

Why doesn't the skater ever regain his potential energy?