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mr Goodwill [35]
2 years ago
8

Paolo says that his bicycle is hard to pedal.Mia looks at the bicycle and tells him he needs to oil the chain.explain why oiling

the chain would help paolo
Physics
1 answer:
laila [671]2 years ago
6 0

Answer:

Less friction

Explanation:

Paolo's bike is too difficult to pedal because there is too much friction in the mechanisms of the bike. To reduce friction, Paolo must oil the chain. This will make the bike run much more smoothly and allow for easier pedalling.

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A child has a mass of 35 kg. The child is running across a fiend and has a speed of 3 m/s. What is the kinetic energy of the chi
Sladkaya [172]

Answer:

Explanation:

Given the following data;

Mass = 35 kg

Velocity = 3 m/s

To find the kinetic energy of the child;

K.E = ½mv²

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3 years ago
What is parkour and how do you do it?
saw5 [17]

Answer:

Parkour is a type of sport or “Hobby”

Explanation:

Parkour is the activity or sport of running through an area, typically in an urban environment.

You can perform parkour by jumping over certain objects, leaping and turning, etc.

Hope this helped!

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2 years ago
Pls help I need this done soon 50 points
lidiya [134]

Answer:

1,3,5

Explanation:

i think maybe dont come at me

6 0
3 years ago
Read 2 more answers
Which is an example of potential en
SIZIF [17.4K]

a yoyo in someones hand is an example of potential energy

8 0
3 years ago
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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
2 years ago
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