To check the correctness of physical equation, v = u + at , Where ‘u’ is the initial velocity, ‘v’ is the final velocity, ‘a’ is the acceleration and ‘t’ is the time in which the change occurs. From (1) and (2) we have [L.H.S.] = [R.H.S.] Hence by principle of homogeneity the given equation is dimensionally correct.
How do you know if an equation is dimensionally correct?
So, both terms have the same dimensions of length. So, we can add both the terms on the right side and the sum will still have the dimensions of length. Since both the sides have the dimensions of length, the equation is dimensionally correct.
How do I check if V v0 at is dimensionally correct?
Note that v and v0 are velocities and that a is an acceleration. Write the dimension of each term. The dimensions of both the sides are the same . Thus, the equation is dimensionally consistent.
How is something dimensionally correct?
In order for an equation to be valid, the dimensions on the left side must match the dimensions on the right side , in which case it is dimensionally correct. ... For an equation to be valid, the dimensions on the left side must match the dimensions on the right side (just like our oranges example.)
How do you use vu?
- u is initial velocity.
- v is the final velocity.
- a is acceleration.
- t is the time period.
Is this dimensionally correct?
An equation in which each term has the same dimensions is said to be dimensionally correct . All equations used in any science should be dimensionally correct. The only time you’ll encounter one which isn’t is if there is an error in the equation.
Is v2 2as dimensionally correct?
And the unit of acceleration is m/s2. Now s = distance and distance is measured in meter or centimeter. Now as we know that the dimension is independent of scaling so the dimension of 2as is [L2T−2] . Hence the given relation is accurate.
Is MGH 1 2mv2 dimensionally correct?
Both sides are dimensionally the same, hence the equations 12mv2 = mgh is dimensionally correct .
Is T 2π √ l g dimensionally correct?
The time period of a simple pendulum is given by T=2π√lg, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct . ... In the above equation, the dimensions of both the LHS and the RHS are the same. This means that the given equation is dimensionally correct.
Which equation of motion is dimensionally incorrect?
$ {u^2} = 2a(gt – 1) $ where $ g $ must be the acceleration due to gravity. Now, from the first principle stated above, option C must be dimensionally incorrect because it has the subtraction of dimensionless constant with a quantity with dimension. Hence, the correct option is option C.
What is dimensionally compatible?
Dimensional compatibility is. achieved when the dimensions of the variables on both sides . of an equation are equal .
What is C in E mc2?
An equation derived by the twentieth-century physicist Albert Einstein, in which E represents units of energy, m represents units of mass, and c2 is the speed of light squared , or multiplied by itself.
How many dimensions are there?
The world as we know it has three dimensions of space —length, width and depth—and one dimension of time. But there’s the mind-bending possibility that many more dimensions exist out there. According to string theory, one of the leading physics model of the last half century, the universe operates with 10 dimensions.
What is the dimension of E mc2?
since the dimension of E is equal to mc^2 ,hence einstein’s equation is dimensionally consistent.
Are all equations dimensionally correct?
Units and Measurement
This statement is wrong. A dimensionally correct equation may or may not be numerically correct. Therefore the equation is dimensionally correct . The angle subtended by an arc of length l, circle of radius r, at the center is given by t/r.
Is first equation of motion is dimensionally correct?
To check the correctness of physical equation, v = u + at, Where ‘u’ is the initial velocity, ‘v’ is the final velocity, ‘a’ is the acceleration and ‘t’ is the time in which the change occurs. From (1) and (2) we have [L.H.S.] = [R.H.S.] Hence by principle of homogeneity the given equation is dimensionally correct.
