Given: - Mass of the body: \( m = 4 \, \text{kg} \) - Forces acting on the body:
\[ \vec{F_1} = 5\hat{i} + 8\hat{j} + 7\hat{k} \] \[ \vec{F_2} = 3\hat{i} - 4\hat{j} - 3\hat{k} \]
The net force acting on the body is given by the vector sum of \(\vec{F_1}\) and \(\vec{F_2}\):
\[ \vec{F_{\text{net}}} = \vec{F_1} + \vec{F_2} \]
Substituting the given values:
\[ \vec{F_{\text{net}}} = (5\hat{i} + 8\hat{j} + 7\hat{k}) + (3\hat{i} - 4\hat{j} - 3\hat{k}) \]
Combining like terms:
\[ \vec{F_{\text{net}}} = (5 + 3)\hat{i} + (8 - 4)\hat{j} + (7 - 3)\hat{k} \] \[ \vec{F_{\text{net}}} = 8\hat{i} + 4\hat{j} + 4\hat{k} \]
The acceleration \(\vec{a}\) is given by Newton’s second law:
\[ \vec{a} = \frac{\vec{F_{\text{net}}}}{m} \]
Substituting the values:
\[ \vec{a} = \frac{1}{4} (8\hat{i} + 4\hat{j} + 4\hat{k}) \] \[ \vec{a} = 2\hat{i} + \hat{j} + \hat{k} \]
The acceleration acting on the body is \( 2\hat{i} + \hat{j} + \hat{k} \).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32