QuestionFind the dimensions of constants $A$ and $B$ in $V(t)=A t^{3}+\frac{B}{t}$ . Options are given for both constants.

Studdy Solution

STEP 1

Assumptions1. The volume V of an object depends on time t according to the equation $V(t)=A t^{3}+\frac{B}{t}$ . . A and B are constants.
3. L and denote dimensions of length and time, respectively.

STEP 2

We are asked to find the dimensions of the constants A and B. The dimensions of a quantity are the powers to which the base quantities must be raised to represent that quantity.

STEP 3

The volume V has dimensions of length cubed, or $^{3}$ .

STEP 4

The first term in the equation for V is $A t^{3}$ . Since the volume V has dimensions of $^{3}$ , and time t has dimensions of, the dimensions of the constant A must be such that when multiplied by $^{3}$ , it gives $^{3}$ .

STEP 5

So, we can write the dimensions of A as $[A] = \frac{[V]}{[t^{3}]}$

STEP 6

Substitute the dimensions of V and t into the equation.
$[A] = \frac{^{3}}{^{3}}$ So, the dimension of the constant A is $\mathrm{}^{3} / \mathrm{T}^{3}$ .

STEP 7

Now, let's find the dimensions of the constant B. The second term in the equation for V is $\frac{B}{t}$ .

STEP 8

The dimensions of the constant B must be such that when divided by, it gives $^{3}$ .

STEP 9

So, we can write the dimensions of B as $[B] = [V] \cdot [t]$

STEP 10

Substitute the dimensions of V and t into the equation.
$[B] = L^{3} \cdot$ So, the dimension of the constant B is $\mathrm{}^{3} \cdot \mathrm{T}$ .

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QuestionFind the dimensions of constants $A$ and $B$ in $V(t)=A t^{3}+\frac{B}{t}$ . Options are given for both constants.

Studdy Solution

STEP 1

Assumptions1. The volume V of an object depends on time t according to the equation $V(t)=A t^{3}+\frac{B}{t}$ . . A and B are constants.
3. L and denote dimensions of length and time, respectively.

STEP 2

We are asked to find the dimensions of the constants A and B. The dimensions of a quantity are the powers to which the base quantities must be raised to represent that quantity.

STEP 3

The volume V has dimensions of length cubed, or $^{3}$ .

STEP 4

The first term in the equation for V is $A t^{3}$ . Since the volume V has dimensions of $^{3}$ , and time t has dimensions of, the dimensions of the constant A must be such that when multiplied by $^{3}$ , it gives $^{3}$ .

STEP 5

So, we can write the dimensions of A as $[A] = \frac{[V]}{[t^{3}]}$

STEP 6

Substitute the dimensions of V and t into the equation.
$[A] = \frac{^{3}}{^{3}}$ So, the dimension of the constant A is $\mathrm{}^{3} / \mathrm{T}^{3}$ .

STEP 7

Now, let's find the dimensions of the constant B. The second term in the equation for V is $\frac{B}{t}$ .

STEP 8

The dimensions of the constant B must be such that when divided by, it gives $^{3}$ .

STEP 9

So, we can write the dimensions of B as $[B] = [V] \cdot [t]$

STEP 10

Substitute the dimensions of V and t into the equation.
$[B] = L^{3} \cdot$ So, the dimension of the constant B is $\mathrm{}^{3} \cdot \mathrm{T}$ .

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QuestionFind the dimensions of constants AAA and BBB in V(t)=At3+BtV(t)=A t^{3}+\frac{B}{t}V(t)=At3+tB​. Options are given for both constants.

Studdy Solution

STEP 1

Assumptions1. The volume V of an object depends on time t according to the equation V(t)=At3+BtV(t)=A t^{3}+\frac{B}{t}V(t)=At3+tB​. . A and B are constants.3. L and denote dimensions of length and time, respectively.

STEP 2

We are asked to find the dimensions of the constants A and B. The dimensions of a quantity are the powers to which the base quantities must be raised to represent that quantity.

STEP 3

The volume V has dimensions of length cubed, or 3^{3}3.

STEP 4

The first term in the equation for V is At3A t^{3}At3. Since the volume V has dimensions of 3^{3}3, and time t has dimensions of, the dimensions of the constant A must be such that when multiplied by 3^{3}3, it gives 3^{3}3.

STEP 5

So, we can write the dimensions of A as[A]=[V][t3][A] = \frac{[V]}{[t^{3}]}[A]=[t3][V]​

STEP 6

Substitute the dimensions of V and t into the equation.[A]=33[A] = \frac{^{3}}{^{3}}[A]=33​So, the dimension of the constant A is 3/T3\mathrm{}^{3} / \mathrm{T}^{3}3/T3.

STEP 7

Now, let's find the dimensions of the constant B. The second term in the equation for V is Bt\frac{B}{t}tB​.

STEP 8

The dimensions of the constant B must be such that when divided by, it gives 3^{3}3.

STEP 9

So, we can write the dimensions of B as[B]=[V]⋅[t][B] = [V] \cdot [t][B]=[V]⋅[t]

STEP 10

Substitute the dimensions of V and t into the equation.[B]=L3⋅[B] = L^{3} \cdot[B]=L3⋅So, the dimension of the constant B is 3⋅T\mathrm{}^{3} \cdot \mathrm{T}3⋅T.

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QuestionFind the dimensions of constants $A$ and $B$ in $V(t)=A t^{3}+\frac{B}{t}$ . Options are given for both constants.

Assumptions1. The volume V of an object depends on time t according to the equation $V(t)=A t^{3}+\frac{B}{t}$ . . A and B are constants.
3. L and denote dimensions of length and time, respectively.

The volume V has dimensions of length cubed, or $^{3}$ .

The first term in the equation for V is $A t^{3}$ . Since the volume V has dimensions of $^{3}$ , and time t has dimensions of, the dimensions of the constant A must be such that when multiplied by $^{3}$ , it gives $^{3}$ .

So, we can write the dimensions of A as $[A] = \frac{[V]}{[t^{3}]}$

Substitute the dimensions of V and t into the equation.
$[A] = \frac{^{3}}{^{3}}$ So, the dimension of the constant A is $\mathrm{}^{3} / \mathrm{T}^{3}$ .

Now, let's find the dimensions of the constant B. The second term in the equation for V is $\frac{B}{t}$ .

The dimensions of the constant B must be such that when divided by, it gives $^{3}$ .

So, we can write the dimensions of B as $[B] = [V] \cdot [t]$

Substitute the dimensions of V and t into the equation.
$[B] = L^{3} \cdot$ So, the dimension of the constant B is $\mathrm{}^{3} \cdot \mathrm{T}$ .