Three blocks $A, B$ and $C$ of masses $4\, kg$,$2\, kg$ and $1\, kg$ respectively,are in contact on a frictionless surface,as shown. If a force of $14\, N$ is applied on the $4\, kg$ block,then the contact force between $A$ and $B$ is .......... $N$.
- A$6$
- B$8$
- C$18$
- D$2$
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Similar Questions
Observe the following statements:
$(I)$ In $\triangle ABC, b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC, \cot \frac{A}{2} = \frac{b+c}{a} \Rightarrow B = 90^{\circ}$
Which of the following is correct?
$(I)$ In $\triangle ABC, b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC, \cot \frac{A}{2} = \frac{b+c}{a} \Rightarrow B = 90^{\circ}$
Which of the following is correct?
The kind of tissue that forms the supportive structure in the pinna is also found in
Two simple pendulums whose lengths are $1\,m$ and $121\,cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will the two be in phase again?
The product $(A)$ is:
The rate of the reaction $2N_2O_5 \to 4NO_2 + O_2$ can be written in three ways:
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:
$-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$\frac{d[NO_2]}{dt} = k'[N_2O_5]$
$\frac{d[O_2]}{dt} = k''[N_2O_5]$
The relationship between $k$ and $k'$ and between $k$ and $k''$ are:
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