Question
A 0.15 kg baseball is moving 50 m/s. What is its kinetic energy?
Answers
Answer
A crane suspends a 1000kg pallet of steel 100 m in the air. g = 9.81 m/s/s.
Answer
Question A large spring, has a spring constant of 10 N/m. It is stretched 2 m. What is the spring p
Answer
What is the kinetic energy of a 2 kg glider moving at 3 m/s?
Answer
On planet easy math where the value of "g" is 10 m/s/s, there is a 2 kg model train is at the top of a 5 meter tall ramp. How much gravity potential energy does the model train have?
Answer
How much energy does a stiff spring with a spring constant of 10 N/m stretched 1 meter?
Answered by
GPT-4o mini
The kinetic energy (KE) of an object can be calculated using the formula:
\[
\text{KE} = \frac{1}{2} mv^2
\]
where \( m \) is the mass of the object in kilograms and \( v \) is the velocity in meters per second.
For a baseball with a mass of 0.15 kg moving at a velocity of 50 m/s, we can substitute the values into the formula:
\[
\text{KE} = \frac{1}{2} \times 0.15 \, \text{kg} \times (50 \, \text{m/s})^2
\]
Calculating this step by step:
1. Calculate \( (50 \, \text{m/s})^2 \):
\[
(50 \, \text{m/s})^2 = 2500 \, \text{m}^2/\text{s}^2
\]
2. Now substitute back into the kinetic energy formula:
\[
\text{KE} = \frac{1}{2} \times 0.15 \times 2500
\]
3. Calculate \( 0.15 \times 2500 \):
\[
0.15 \times 2500 = 375
\]
4. Now, compute \( \frac{1}{2} \times 375 \):
\[
\text{KE} = \frac{375}{2} = 187.5 \, \text{J}
\]
Thus, the kinetic energy of the baseball is \( 187.5 \, \text{J} \).
\[
\text{KE} = \frac{1}{2} mv^2
\]
where \( m \) is the mass of the object in kilograms and \( v \) is the velocity in meters per second.
For a baseball with a mass of 0.15 kg moving at a velocity of 50 m/s, we can substitute the values into the formula:
\[
\text{KE} = \frac{1}{2} \times 0.15 \, \text{kg} \times (50 \, \text{m/s})^2
\]
Calculating this step by step:
1. Calculate \( (50 \, \text{m/s})^2 \):
\[
(50 \, \text{m/s})^2 = 2500 \, \text{m}^2/\text{s}^2
\]
2. Now substitute back into the kinetic energy formula:
\[
\text{KE} = \frac{1}{2} \times 0.15 \times 2500
\]
3. Calculate \( 0.15 \times 2500 \):
\[
0.15 \times 2500 = 375
\]
4. Now, compute \( \frac{1}{2} \times 375 \):
\[
\text{KE} = \frac{375}{2} = 187.5 \, \text{J}
\]
Thus, the kinetic energy of the baseball is \( 187.5 \, \text{J} \).
Answered by
GPT-4o mini
To find the gravitational potential energy (PE) of the suspended pallet of steel, we can use the formula:
\[
\text{PE} = mgh
\]
where:
- \( m \) is the mass of the object (in kilograms),
- \( g \) is the acceleration due to gravity (in meters per second squared),
- \( h \) is the height above the reference point (in meters).
In this case:
- \( m = 1000 \, \text{kg} \) (mass of the pallet),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 100 \, \text{m} \) (height above ground).
Now we can substitute these values into the potential energy formula:
\[
\text{PE} = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 100 \, \text{m}
\]
Calculating this step by step:
1. Calculate \( m \times g \):
\[
1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N}
\]
2. Now, calculate the potential energy:
\[
\text{PE} = 9810 \, \text{N} \times 100 \, \text{m} = 981000 \, \text{J}
\]
Therefore, the gravitational potential energy of the suspended pallet of steel is \( 981000 \, \text{J} \) or \( 981 \, \text{kJ} \).
\[
\text{PE} = mgh
\]
where:
- \( m \) is the mass of the object (in kilograms),
- \( g \) is the acceleration due to gravity (in meters per second squared),
- \( h \) is the height above the reference point (in meters).
In this case:
- \( m = 1000 \, \text{kg} \) (mass of the pallet),
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 100 \, \text{m} \) (height above ground).
Now we can substitute these values into the potential energy formula:
\[
\text{PE} = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 100 \, \text{m}
\]
Calculating this step by step:
1. Calculate \( m \times g \):
\[
1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N}
\]
2. Now, calculate the potential energy:
\[
\text{PE} = 9810 \, \text{N} \times 100 \, \text{m} = 981000 \, \text{J}
\]
Therefore, the gravitational potential energy of the suspended pallet of steel is \( 981000 \, \text{J} \) or \( 981 \, \text{kJ} \).
Answered by
GPT-4o mini
To find the potential energy stored in a spring that is stretched (or compressed), you can use the formula for spring potential energy (also known as elastic potential energy):
\[
PE = \frac{1}{2} k x^2
\]
where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant (in N/m),
- \( x \) is the displacement from the equilibrium position (in meters).
Given:
- \( k = 10 \, \text{N/m} \)
- \( x = 2 \, \text{m} \)
Now, substituting the values into the formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times (2 \, \text{m})^2
\]
Calculating it step by step:
1. Calculate \( (2 \, \text{m})^2 \):
\[
(2 \, \text{m})^2 = 4 \, \text{m}^2
\]
2. Now substitute this back into the potential energy formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times 4 \, \text{m}^2
\]
3. Calculate \( 10 \, \text{N/m} \times 4 \, \text{m}^2 \):
\[
10 \times 4 = 40
\]
4. Now calculate \( \frac{1}{2} \times 40 \):
\[
PE = \frac{40}{2} = 20 \, \text{J}
\]
Therefore, the potential energy stored in the spring when it is stretched 2 meters is **20 Joules**.
\[
PE = \frac{1}{2} k x^2
\]
where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant (in N/m),
- \( x \) is the displacement from the equilibrium position (in meters).
Given:
- \( k = 10 \, \text{N/m} \)
- \( x = 2 \, \text{m} \)
Now, substituting the values into the formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times (2 \, \text{m})^2
\]
Calculating it step by step:
1. Calculate \( (2 \, \text{m})^2 \):
\[
(2 \, \text{m})^2 = 4 \, \text{m}^2
\]
2. Now substitute this back into the potential energy formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times 4 \, \text{m}^2
\]
3. Calculate \( 10 \, \text{N/m} \times 4 \, \text{m}^2 \):
\[
10 \times 4 = 40
\]
4. Now calculate \( \frac{1}{2} \times 40 \):
\[
PE = \frac{40}{2} = 20 \, \text{J}
\]
Therefore, the potential energy stored in the spring when it is stretched 2 meters is **20 Joules**.
Answered by
GPT-4o mini
To find the kinetic energy (KE) of a glider, you can use the kinetic energy formula:
\[
\text{KE} = \frac{1}{2} mv^2
\]
where:
- \( m \) is the mass of the object (in kilograms),
- \( v \) is the velocity of the object (in meters per second).
Given:
- \( m = 2 \, \text{kg} \)
- \( v = 3 \, \text{m/s} \)
We can substitute these values into the formula:
\[
\text{KE} = \frac{1}{2} \times 2 \, \text{kg} \times (3 \, \text{m/s})^2
\]
Calculating it step by step:
1. Calculate \( (3 \, \text{m/s})^2 \):
\[
(3 \, \text{m/s})^2 = 9 \, \text{m}^2/\text{s}^2
\]
2. Now substitute this back into the kinetic energy formula:
\[
\text{KE} = \frac{1}{2} \times 2 \, \text{kg} \times 9 \, \text{m}^2/\text{s}^2
\]
3. Calculate \( 2 \, \text{kg} \times 9 \, \text{m}^2/\text{s}^2 \):
\[
2 \times 9 = 18
\]
4. Now calculate \( \frac{1}{2} \times 18 \):
\[
\text{KE} = \frac{18}{2} = 9 \, \text{J}
\]
Therefore, the kinetic energy of the 2 kg glider moving at 3 m/s is **9 Joules**.
\[
\text{KE} = \frac{1}{2} mv^2
\]
where:
- \( m \) is the mass of the object (in kilograms),
- \( v \) is the velocity of the object (in meters per second).
Given:
- \( m = 2 \, \text{kg} \)
- \( v = 3 \, \text{m/s} \)
We can substitute these values into the formula:
\[
\text{KE} = \frac{1}{2} \times 2 \, \text{kg} \times (3 \, \text{m/s})^2
\]
Calculating it step by step:
1. Calculate \( (3 \, \text{m/s})^2 \):
\[
(3 \, \text{m/s})^2 = 9 \, \text{m}^2/\text{s}^2
\]
2. Now substitute this back into the kinetic energy formula:
\[
\text{KE} = \frac{1}{2} \times 2 \, \text{kg} \times 9 \, \text{m}^2/\text{s}^2
\]
3. Calculate \( 2 \, \text{kg} \times 9 \, \text{m}^2/\text{s}^2 \):
\[
2 \times 9 = 18
\]
4. Now calculate \( \frac{1}{2} \times 18 \):
\[
\text{KE} = \frac{18}{2} = 9 \, \text{J}
\]
Therefore, the kinetic energy of the 2 kg glider moving at 3 m/s is **9 Joules**.
Answered by
GPT-4o mini
To calculate the gravitational potential energy (PE) of the model train at the top of the ramp, you can use the formula:
\[
PE = mgh
\]
where:
- \( PE \) is the gravitational potential energy (in joules),
- \( m \) is the mass of the object (in kilograms),
- \( g \) is the acceleration due to gravity (in meters per second squared),
- \( h \) is the height above the reference point (in meters).
Given:
- \( m = 2 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 5 \, \text{m} \)
Now, substituting these values into the formula:
\[
PE = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5 \, \text{m}
\]
Calculating it step by step:
1. Calculate \( 10 \, \text{m/s}^2 \times 5 \, \text{m} \):
\[
10 \times 5 = 50 \, \text{N}
\]
2. Now calculate \( 2 \, \text{kg} \times 50 \, \text{N} \):
\[
2 \, \text{kg} \times 50 \, \text{N} = 100 \, \text{J}
\]
Therefore, the gravitational potential energy of the model train at the top of the 5-meter ramp is **100 Joules**.
\[
PE = mgh
\]
where:
- \( PE \) is the gravitational potential energy (in joules),
- \( m \) is the mass of the object (in kilograms),
- \( g \) is the acceleration due to gravity (in meters per second squared),
- \( h \) is the height above the reference point (in meters).
Given:
- \( m = 2 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 5 \, \text{m} \)
Now, substituting these values into the formula:
\[
PE = 2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5 \, \text{m}
\]
Calculating it step by step:
1. Calculate \( 10 \, \text{m/s}^2 \times 5 \, \text{m} \):
\[
10 \times 5 = 50 \, \text{N}
\]
2. Now calculate \( 2 \, \text{kg} \times 50 \, \text{N} \):
\[
2 \, \text{kg} \times 50 \, \text{N} = 100 \, \text{J}
\]
Therefore, the gravitational potential energy of the model train at the top of the 5-meter ramp is **100 Joules**.
Answered by
GPT-4o mini
To calculate the potential energy stored in a spring that is stretched (or compressed), you can use the formula for spring potential energy (also known as elastic potential energy):
\[
PE = \frac{1}{2} k x^2
\]
where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant (in N/m),
- \( x \) is the displacement from the equilibrium position (in meters).
Given:
- \( k = 10 \, \text{N/m} \)
- \( x = 1 \, \text{m} \)
Now substituting these values into the formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times (1 \, \text{m})^2
\]
Calculating it step by step:
1. Calculate \( (1 \, \text{m})^2 \):
\[
(1 \, \text{m})^2 = 1 \, \text{m}^2
\]
2. Substitute this back into the potential energy formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times 1 \, \text{m}^2
\]
3. Calculate \( 10 \, \text{N/m} \times 1 \, \text{m}^2 \):
\[
10 \times 1 = 10
\]
4. Now calculate \( \frac{1}{2} \times 10 \):
\[
PE = \frac{10}{2} = 5 \, \text{J}
\]
Therefore, the energy stored in the stiff spring when it is stretched 1 meter is **5 Joules**.
\[
PE = \frac{1}{2} k x^2
\]
where:
- \( PE \) is the potential energy,
- \( k \) is the spring constant (in N/m),
- \( x \) is the displacement from the equilibrium position (in meters).
Given:
- \( k = 10 \, \text{N/m} \)
- \( x = 1 \, \text{m} \)
Now substituting these values into the formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times (1 \, \text{m})^2
\]
Calculating it step by step:
1. Calculate \( (1 \, \text{m})^2 \):
\[
(1 \, \text{m})^2 = 1 \, \text{m}^2
\]
2. Substitute this back into the potential energy formula:
\[
PE = \frac{1}{2} \times 10 \, \text{N/m} \times 1 \, \text{m}^2
\]
3. Calculate \( 10 \, \text{N/m} \times 1 \, \text{m}^2 \):
\[
10 \times 1 = 10
\]
4. Now calculate \( \frac{1}{2} \times 10 \):
\[
PE = \frac{10}{2} = 5 \, \text{J}
\]
Therefore, the energy stored in the stiff spring when it is stretched 1 meter is **5 Joules**.
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