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3 years ago

Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.

2 answers:

6 0

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

Mnenie [13.5K]3 years ago

3 0

Question

Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.

Answer

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

So it starts as potential engry sence they are staying still but then been they strt moveing by pedleing upnthe hill it turns/or changes in kentic energy sence we are gping up the when i go bake down i don't use as much force or have to push because the gravity is pulling me down thats when it turns into gravitational energy.

Hope this helps :)

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2 years ago

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B

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Explanation:

Let $\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}$ and $\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}$

The sum of the two vectors is

$\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}$

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The difference between the two vectors can be written as

$\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}$

$= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}$

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3 years ago

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2 years ago

Read 2 more answers

Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r

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The force of gravity between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
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r is their separation

In this problem, the mass of the object is $m_1=0.60 kg$ , while the Earth's mass is $m_2=5.97 \cdot 10^{24} kg$ . Their separation is $r=1.3 \cdot 10^7 m$ , therefore the gravitational force exerted on the object is
$F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N$

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2 years ago