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2 years ago

5

Why the bicycle stops when the cyclist stops pedaling?

2 answers:

Sladkaya [172]2 years ago

5 0

There is no kintentic energy being giving to the bike through the action of pedaling therefore the bike will stop moving

mr_godi [17]2 years ago

3 0

Answer:

The bicycle slows down because of the frictional force acting on the tyres of the bicycle. When this frictional force overcomes the force applied by paddling, the bicycle stops.

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Explain where most of the mass of an atom is located. Also, explain why some particles that make up the atom do not contribute m

Naddika [18.5K]

Within an atom, there are three elementary particles: the proton, neutron, and electron. Most of the mass of an atom is situated within the nucleus, which is the central part of the atom. It is made up of protons and neutrons, which are the heaviest subatomic particles. The electrons within the atom, orbit around the nucleus at a very far distance. Electrons are also a part of the lightest group of subatomic particles called leptons. That is why these electrons don't contribute much to the majority of an atoms mass. They are very light and they orbit at very far distances.

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2 years ago

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Que es electromagnetico?

adoni [48]

Se llaman cuerpos eléctricos.

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2 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

2 years ago

Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11

tiny-mole [99]

Answer:

The line charge density is $1.59\times10^{-4}\ C/m$

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

$V=EA$

$V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2$

$\lambda=\dfrac{V\times2\epsilon_{0}}{r}$

Where, r = radius

V = potential difference

Put the value into the formula

$\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}$

$\lambda=1.59\times10^{-4}\ C/m$

Hence, The line charge density is $1.59\times10^{-4}\ C/m$

4 0

2 years ago

There is T__ as much voltage. This<br> produces twice the current

Lera25 [3.4K]

Twice as much voltage

3 0

2 years ago