e the illustration below. The formula V= m + M - (2gh)!/2 m es the velocity (in ft/sec) of a bullet with weight m fired into a block with weight M, that raises the height of the block h feet after the collision. The letter ocity of the bullet to the nearest ft/sec, if m = 0.0624, M = 6, and h = 0.9. represents the constant, 32. Find the ft/sec m lb I hft M lb
Quadratic Equation
When it comes to the concept of polynomial equations, quadratic equations can be said to be a special case. What does solving a quadratic equation mean? We will understand the quadratics and their types once we are familiar with the polynomial equations and their types.
Demand and Supply Function
The concept of demand and supply is important for various factors. One of them is studying and evaluating the condition of an economy within a given period of time. The analysis or evaluation of the demand side factors are important for the suppliers to understand the consumer behavior. The evaluation of supply side factors is important for the consumers in order to understand that what kind of combination of goods or what kind of goods and services he or she should consume in order to maximize his utility and minimize the cost. Therefore, in microeconomics both of these concepts are extremely important in order to have an idea that what exactly is going on in the economy.
![### Energy and Momentum in Collisions
#### Problem Statement:
Given a scenario where a bullet with weight \( m \) is fired into a block with weight \( M \), causing the block to rise to a height \( h \) after the collision, we can determine the velocity of the bullet using the formula:
\[ v = \frac{m + M}{m} \sqrt{2gh} \]
where:
- \( v \) is the velocity of the bullet in ft/sec,
- \( g \), representing the gravitational constant, is equal to 32 ft/sec²,
- \( m \) is the weight of the bullet in lb,
- \( M \) is the weight of the block in lb,
- \( h \) is the height in feet that the block rises after the collision.
#### Given:
- \( m = 0.0624 \) lb,
- \( M = 6 \) lb,
- \( h = 0.9 \) ft.
#### Objective:
Find the velocity \( v \) of the bullet to the nearest ft/sec.
##### Calculation:
\[ v = \frac{m + M}{m} \sqrt{2gh} \]
Let's plug in the given values:
\[ v = \frac{0.0624 + 6}{0.0624} \sqrt{2 \times 32 \times 0.9} \]
We can break down this calculation into several steps:
1. Sum \( m \) and \( M \):
\[ 0.0624 + 6 = 6.0624 \]
2. Divide by \( m \):
\[ \frac{6.0624}{0.0624} = 97.2 \]
3. Calculate \( 2gh \):
\[ 2 \times 32 \times 0.9 = 57.6 \]
4. Take the square root of \( 2gh \):
\[ \sqrt{57.6} \approx 7.583 \]
5. Multiply these values together to find \( v \):
\[ v = 97.2 \times 7.583 \approx 737.08 \]
So, rounding to the nearest ft/sec, the velocity \( v \) of the bullet is:
\[ v = \boxed{737} \text{ ft/sec} \]
#### Diagram Analysis:
The provided diagram illustrates:
- A setup with a bullet \( m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c8e8955-7b00-44a9-a272-a38b187e6aa8%2Fe04769bb-cec9-4cfb-aebd-17ac8fd9cda2%2F0y29tq_processed.png&w=3840&q=75)
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