Questions   
Sort: 
Apr 2, 2023
 #1
avatar+50 
+1

From https://brainly.com/question/30733513

Duplicate of https://web2.0calc.com/questions/counting_66045#r1

Dupilcate of https://web2.0calc.com/questions/counting_85335#r1

 

Using combination, the number of ways of placing 8 counters in the square is 12 ways and the number of ways of placing 12 counters in the square is 8 ways

What is the number of ways of placing 8 counters in the square

a) To place 8 counters in the 4 x 4 grid, with exactly two counters in each row, we can first choose the two columns in the first row that will have counters. Using combination, there are {4 C 2} = 6 ways to do this. Then, we choose the two columns in the second row that will have counters, but we need to make sure that these columns are not the same as the ones chosen in the first row. There are only two columns left to choose from, so there are only two ways to do this. Similarly, we choose the two columns in the third row that will have counters, making sure that they are different from the columns chosen in the first two rows. There is only one column left to choose from, so there is only one way to do this. Finally, the columns for the fourth row are uniquely determined. Therefore, the total number of ways to place 8 counters in the grid with exactly two counters in each row is:

6 * 2 = 12

(b) To place 12 counters in the 4 x 4 grid, with exactly three counters in each column, we can start by choosing the three rows that will have counters in the first column. Using combination, there are {4 C3} = 4 ways to do this. Then, we choose the three rows that will have counters in the second column, but we need to make sure that these rows are not the same as the ones chosen for the first column. There are only two rows left to choose from, so there are only two ways to do this. Similarly, we choose the three rows that will have counters in the third column, making sure that they are different from the rows chosen for the first two columns. There is only one row left to choose from, so there is only one way to do this. Finally, the rows for the fourth column are uniquely determined. Therefore, the total number of ways to place 12 counters in the grid with exactly three counters in each column is:

4 * 2 = 8

Thus, there are 12 ways to place 8 counters in the grid with exactly two counters in each row, and 8 ways to place 12 counters in the grid with exactly three counters in each column.

Apr 2, 2023
 #1
avatar
0

We can use the principle of work and energy to solve this problem. The work done by the applied force is equal to the change in the sled's kinetic energy:

�applied=Δ�Wapplied​=ΔK

where $W_\text{applied}$ is the work done by the applied force, and $\Delta K$ is the change in kinetic energy. The work done by the applied force is given by:

�applied=�applied�cos⁡�Wapplied​=Fapplied​dcosθ

where $d$ is the distance traveled by Amara and her sled. We can calculate $d$ using the formula for the distance traveled with constant acceleration:

�=12��2d=21​at2

where $a$ is the acceleration and $t$ is the time. The acceleration of Amara and her sled is given by:

�=�appliedcos⁡�−�friction�a=mFapplied​cosθ−ffriction​​

where $f_\text{friction}$ is the force of friction and $m$ is the mass of Amara and her sled. The force of friction is given by:

�friction=��ffriction​=μN

where $N$ is the normal force, which is equal to the weight of Amara and her sled:

�=��N=mg

where $g$ is the acceleration due to gravity. Putting it all together, we have:

\begin{align*} d &= \frac{1}{2}at^2 \ &= \frac{1}{2} \frac{F_\text{applied} \cos\theta - f_\text{friction}}{m} t^2 \ &= \frac{1}{2} \frac{F_\text{applied} \cos\theta - \mu mg}{m} t^2 \ \end{align*}

The final speed of Amara and her sled is given by:

��2=��2+2�Δ�vf2​=vi2​+2aΔx

where $v_i$ is the initial velocity (which is zero), $\Delta x$ is the distance traveled, and $v_f$ is the final velocity. Solving for $\Delta x$, we get:

Δ�=��22�Δx=2avf2​​

Putting it all together, we have:

\begin{align*} W_\text{applied} &= \Delta K \ F_\text{applied} d \cos\theta &= \frac{1}{2} m v_f^2 \ F_\text{applied} \frac{1}{2} \frac{F_\text{applied} \cos\theta - \mu mg}{m} t^2 \cos\theta &= \frac{1}{2} m v_f^2 \ F_\text{applied} \frac{1}{2} \left(F_\text{applied} \cos^2\theta - \mu mg \cos\theta\right) t^2 &= \frac{1}{2} m v_f^2 \ F_\text{applied} &= \frac{m v_f^2}{t^2 \left(\frac{1}{2} \cos^2\theta - \frac{\mu g}{2}\cos\theta\right)} \end{align*}

Substituting the given values, we get:

F_{applied} = \frac{(50.5\text{ m/s})(5.1 \text{ s}) = 9.9 \text{ N}.

Apr 2, 2023
Apr 1, 2023
Mar 31, 2023

0 Online Users