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).
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
3 answers
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.
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).
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).