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creativ13 [48]
3 years ago
8

How do you know when to use each equations of motion while solving numericals? Please help as soon as possible .

Physics
1 answer:
posledela3 years ago
8 0

Answer:

You must catch the known-variables and the unknown-variables from the question to use the right equation.

Explanation:

Whenever you are solving a numerical, follow these steps:

1.    From the asked question, jot down the given variables like initial velocity, total time, or acceleration etc.

2.    After that, write the variable which is asked and put a question mark next to it.

3.    Now that you have the given information and asked information in terms of variables, you should see which equation has all those variables. It might be helpful to jot down all three equations of motion on a side of paper. Once you have all three equations in front of you, you can easily spot the equation which has all the known and unknown variables in it.

Note:    Sometimes, you might have to use more than one equations to get to the asked variable. You can think of that as two question instead of one, where you have to find out two variable instead of one.

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A person who weighs 150 pounds on Earth would weigh ____ pounds on the moon.
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A skateboarder with a mass of 60 kg moves with a force of 20 N. What is her acceleration?
Zanzabum

Explanation:

Solution,

  • Mass(m)= 60 kg
  • Force (F)= 20 N
  • Acceleration (a)= ?

We know that,

  • F=ma
  • a=F/m
  • a=20/60
  • a=0.333 m/s²

So, her acceleration is 0.333 m/s².

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2 years ago
A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.
Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

d=30,000 pc =9.26\cdot 10^{20} m

The speed of the cosmic ray is

v=0.98 c

where

c=3.0 \cdot 10^8 m/s is the speed of light. Substituting,

v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s

Converting into years,

T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}

And substituting v = 0.98c, we find:

T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years

3 0
3 years ago
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