Asked by unknown 2.0
A student wants to model a system where an orange of mass m is tossed into the air at an initial velocity
v from an initial height h. Which equation best expresses the total initial energy E of the system?
answers:
E = mgh + 1/2mv^2
E = mgh + -1/2mv^2
E = mgh 1/2 mv^2
E = mgh
v from an initial height h. Which equation best expresses the total initial energy E of the system?
answers:
E = mgh + 1/2mv^2
E = mgh + -1/2mv^2
E = mgh 1/2 mv^2
E = mgh
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To calculate the total initial energy \( E \) of the system when the orange is tossed into the air, we need to consider both the gravitational potential energy and the kinetic energy of the orange.
1. The gravitational potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]
2. The kinetic energy (KE) due to the initial velocity \( v \) is given by:
\[ KE = \frac{1}{2}mv^2 \]
Therefore, the total initial energy \( E \) of the system is the sum of the potential energy and the kinetic energy:
\[ E = mgh + \frac{1}{2}mv^2 \]
So the correct equation that expresses the total initial energy \( E \) of the system is:
\[ E = mgh + \frac{1}{2}mv^2 \]
This matches the first answer choice.
1. The gravitational potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]
2. The kinetic energy (KE) due to the initial velocity \( v \) is given by:
\[ KE = \frac{1}{2}mv^2 \]
Therefore, the total initial energy \( E \) of the system is the sum of the potential energy and the kinetic energy:
\[ E = mgh + \frac{1}{2}mv^2 \]
So the correct equation that expresses the total initial energy \( E \) of the system is:
\[ E = mgh + \frac{1}{2}mv^2 \]
This matches the first answer choice.
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