Show that the expression where v represents speed, a acceleration, and t an instant of time, is dimensionally correct.
b .what is dimension of t in the following equation
To show that the expression is dimensionally correct, we need to ensure that all the terms have the same dimensions.
The dimensions of speed (v) are distance divided by time (L/T).
The dimensions of acceleration (a) are distance divided by time squared (L/T^2).
The dimensions of time (t) are time (T).
So the dimensions of v * t - 1/2 * a * t^2 are:
(v * t) - (1/2 * a * t^2) = (L/T * T) - (L/T^2 * T^2) = (L/T * T) - (L/T^2 * T^2) = L - L = 0
Since the dimensions of the expression are zero, we can conclude that the expression is dimensionally correct.
b. The dimension of t in the equation v * t - 1/2 * a * t^2 is time (T).
a.Show that the expression V=at where v represents speed, a acceleration, and t an instant of time, is dimensionally correct.
b.what is dimension of t in the following equation v=u-1/2gt^2.
a) To show that the expression V = at is dimensionally correct, we need to ensure that both sides of the equation have the same dimensions.
The dimensions of speed (V or v) are distance divided by time (L/T).
The dimensions of acceleration (a) are distance divided by time squared (L/T^2).
The dimensions of time (t) are time (T).
On the left side of the equation, we have V, which has dimensions of L/T.
On the right side of the equation, we have a * t, which has dimensions of (L/T^2) * T = L/T.
Since the dimensions on both sides of the equation are L/T, we can conclude that the expression V = at is dimensionally correct.
b) In the equation v = u - (1/2)gt^2, where v represents speed, u represents initial velocity, g represents acceleration due to gravity, and t represents time:
The dimensions of speed (v) are distance divided by time (L/T).
The dimensions of initial velocity (u) are distance divided by time (L/T).
The dimensions of acceleration due to gravity (g) are distance divided by time squared (L/T^2).
The dimensions of time (t) are time (T).
On the left side of the equation, we have v, which has dimensions of L/T.
On the right side of the equation, we have u - (1/2)gt^2. Let's break it down:
- u has dimensions of L/T.
- (1/2)gt^2 has dimensions of (L/T^2) * T^2 = L.
So overall, the dimensions on the right side of the equation are L - L = 0.
Since the dimensions on both sides of the equation are L/T, we can conclude that the dimension of t in the equation v = u - (1/2)gt^2 is time (T).
To show that an expression is dimensionally correct, we need to ensure that the dimensions of all the terms on both sides of the equation are consistent.
The given expression is:
[v] = [a] * [t]
Here, [v] represents the dimension of velocity or speed, [a] represents the dimension of acceleration, and [t] represents the dimension of time.
The dimension of velocity, [v], can be expressed as length divided by time ([L]/[T]).
The dimension of acceleration, [a], can be expressed as change in velocity divided by time, which is (length divided by time) divided by time, or [L]/[T^2].
Now, let's substitute these dimensions into the expression:
[L]/[T] = ([L]/[T^2]) * [T]
Simplifying the expression, we cancel out the time units:
[L] = [L]
This equation shows that the dimensions on both sides are consistent, which means the expression is dimensionally correct.
b. The dimension of time, [t], in the given equation is time ([T]).