College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Question
Help please
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**Question:**
How high will a winch motor with a power of 483 Watts lift an 18 kg crate straight up against gravity in 29 seconds?
**Your Answer:**
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*Explanation for Educational Purposes:*
To determine the height the crate will be lifted, you can use the formula for power:
\[ \text{Power (P)} = \frac{\text{Work (W)}}{\text{Time (t)}} \]
Where Work (W) is defined as:
\[ W = \text{Force (F)} \times \text{Distance (d)} \]
The force here is equal to the weight of the crate, which is mass (m) times gravity (g):
\[ F = m \times g \]
Substituting these into the power formula gives:
\[ P = \frac{m \times g \times d}{t} \]
Rearrange this to solve for distance (d):
\[ d = \frac{P \times t}{m \times g} \]
Inserting the values:
- Power (P) = 483 Watts
- Time (t) = 29 seconds
- Mass (m) = 18 kg
- Gravity (g) = 9.81 m/s² (approximately)
\[ d = \frac{483 \times 29}{18 \times 9.81} \]
Calculate this value to find how high the crate will be lifted.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00114bb1-0365-4dfa-83cb-10b10c9a61b0%2Fb406ecf9-3c17-471e-a785-950676b9b1f9%2F565gcky_processed.jpeg&w=3840&q=75)
Transcribed Image Text:---
**Question:**
How high will a winch motor with a power of 483 Watts lift an 18 kg crate straight up against gravity in 29 seconds?
**Your Answer:**
---
*Explanation for Educational Purposes:*
To determine the height the crate will be lifted, you can use the formula for power:
\[ \text{Power (P)} = \frac{\text{Work (W)}}{\text{Time (t)}} \]
Where Work (W) is defined as:
\[ W = \text{Force (F)} \times \text{Distance (d)} \]
The force here is equal to the weight of the crate, which is mass (m) times gravity (g):
\[ F = m \times g \]
Substituting these into the power formula gives:
\[ P = \frac{m \times g \times d}{t} \]
Rearrange this to solve for distance (d):
\[ d = \frac{P \times t}{m \times g} \]
Inserting the values:
- Power (P) = 483 Watts
- Time (t) = 29 seconds
- Mass (m) = 18 kg
- Gravity (g) = 9.81 m/s² (approximately)
\[ d = \frac{483 \times 29}{18 \times 9.81} \]
Calculate this value to find how high the crate will be lifted.
Expert Solution
Step 1
Given data:
Power (P) = 483 Watts
Mass (m) = 18 kg
Time (t) = 29 seconds
Required:
The height (h)
Step by step
Solved in 2 steps
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