Solving a Physics Problem: Finding the Right Units

  • Thread starter Thread starter ProBasket
  • Start date Start date
  • Tags Tags
    Physics Units
AI Thread Summary
In solving physics problems, it's crucial to determine the correct units for the final answer through dimensional analysis. The example provided shows that when calculating the expression \sqrt{\frac{940N/m}{0.038kg}}*(0.25m), the resulting units simplify to \sqrt{N}*m. The discussion emphasizes that while some units may not cancel out, maintaining consistent and correct units ensures the accuracy of the final answer. Understanding the base units, such as Newtons (N) as kg m s^-2, aids in this process. Ultimately, careful attention to units is essential for successful problem-solving in physics.
ProBasket
140
0
i just need help with finding the units to a physics problem at the end. some examples...


\sqrt{\frac{940N/m}{0.038kg}}*(0.25m)

it comes out to 39m/s.

i know that since it's velocity, it should always be m/s. but there are some question that i don't really know what the final units should be.

without the numbers...
\sqrt{\frac{N/m}{kg}}*(m)
it doesn't seem like anything can be canceled out because the units are being sqrtrooted.
 
Physics news on Phys.org
Put Newtons in base units, N = kg m s^-2 and you'll see everything inside the square root cancels out, besides the s^-2.
 



When solving a physics problem, it is important to pay attention to the units involved in the given values and the final answer. Units act as a guide to ensure that the calculations are done correctly and the final answer is in the correct form. In the given example, the units for force are Newtons (N) and for mass are kilograms (kg). To find the units for the final answer, we can use the method of dimensional analysis.

First, let's look at the given expression: \sqrt{\frac{940N/m}{0.038kg}}*(0.25m). We can see that the numerator has units of N/m (force per unit length) and the denominator has units of kg (mass). When we take the square root, the units of N/m and kg will also be square rooted. This means that the final units for the expression will be \sqrt{\frac{N}{m}}*\sqrt{kg}*m. To simplify this, we can rewrite it as \sqrt{N}*m.

Now, let's consider the example without numbers: \sqrt{\frac{N/m}{kg}}*(m). Here, we can see that the units for force are still N/m and for mass are kg. When we take the square root, the units will be \sqrt{\frac{N}{m}}*\sqrt{kg}. This cannot be simplified any further as the units are being square rooted. Therefore, the final units for this expression will be \sqrt{N}*m.

In conclusion, when solving a physics problem, it is important to pay attention to the units involved and use dimensional analysis to determine the units for the final answer. In some cases, the units may not cancel out and the final answer will have a combination of units. However, as long as the units are correct and consistent, the final answer will be accurate.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up will act downwards and maximum static friction will act downwards Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top