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Answer :
Sure, let's solve this step-by-step.
We are given the formula to find potential energy (PE):
[tex]\[ \text{PE} = m \times g \times h \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the bicycle
- [tex]\( g \)[/tex] is the acceleration due to gravity
- [tex]\( h \)[/tex] is the height of the hill
Let's plug in the values provided:
- [tex]\( m = 25 \)[/tex] kg (mass of the bicycle)
- [tex]\( g = 9.8 \)[/tex] m/s[tex]\(^2\)[/tex] (acceleration due to gravity)
- [tex]\( h = 3 \)[/tex] m (height of the hill)
Now, we substitute these values into the formula:
[tex]\[ \text{PE} = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
1. First, multiply the mass ([tex]\( 25 \)[/tex] kg) by the acceleration due to gravity ([tex]\( 9.8 \)[/tex] m/s[tex]\(^2\)[/tex]):
[tex]\[ 25 \times 9.8 = 245 \][/tex]
2. Next, multiply the result by the height ([tex]\( 3 \)[/tex] m):
[tex]\[ 245 \times 3 = 735 \][/tex]
Therefore, the potential energy of the bicycle resting at the top of the hill is [tex]\( 735 \)[/tex] Joules.
So the correct answer is:
[tex]\[ 735 \, \text{J} \][/tex]
We are given the formula to find potential energy (PE):
[tex]\[ \text{PE} = m \times g \times h \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the bicycle
- [tex]\( g \)[/tex] is the acceleration due to gravity
- [tex]\( h \)[/tex] is the height of the hill
Let's plug in the values provided:
- [tex]\( m = 25 \)[/tex] kg (mass of the bicycle)
- [tex]\( g = 9.8 \)[/tex] m/s[tex]\(^2\)[/tex] (acceleration due to gravity)
- [tex]\( h = 3 \)[/tex] m (height of the hill)
Now, we substitute these values into the formula:
[tex]\[ \text{PE} = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
1. First, multiply the mass ([tex]\( 25 \)[/tex] kg) by the acceleration due to gravity ([tex]\( 9.8 \)[/tex] m/s[tex]\(^2\)[/tex]):
[tex]\[ 25 \times 9.8 = 245 \][/tex]
2. Next, multiply the result by the height ([tex]\( 3 \)[/tex] m):
[tex]\[ 245 \times 3 = 735 \][/tex]
Therefore, the potential energy of the bicycle resting at the top of the hill is [tex]\( 735 \)[/tex] Joules.
So the correct answer is:
[tex]\[ 735 \, \text{J} \][/tex]
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