Miswer

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

4 years ago

The sampling rate of an ADC is 8.1 kHz. What will be an appropriate cut-off frequency (break frequency) for the anti-aliasing fi

KiRa [710]

Answer:

4000 Hz

Explanation:

An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.

The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.

Sampling rate = 2(max input frequency)

From the relation we can find out the cut-off frequency for the anti-aliasing filter.

max input frequency = sampling rate/2

max input frequency = 8100/2 = 4050 Hz

Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.

3 0

3 years ago

Which of the following is not a mineral? A. Gold B. Natural diamond C. Solid ice D. Coal

IgorC [24]

d. coal cause yah coal

3 0

3 years ago

Read 2 more answers

A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by

densk [106]

Answer:

The maximun distance is $z_1 = z_2 = 0.0138m$

Explanation:

From the question we are told that

The wavelength are $\lambda _ 1 = 540nm (green) = 540 *10^{-9}m$

$\lambda_2 = 450nm(blue) = 450 *10^{-9}m$

The distance of seperation of the two slit is $d = 0.180mm = 0.180 *10^{-3}m$

The distance from the screen is $D = 1.53m$

Generally the distance of the bright fringe to the center of the screen is mathematically represented as

$z = \frac{m \lambda D}{d}$

Where m is the order of the fringe

For the first wavelength we have

$z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}$

$z_1=0.00459m_1 m$

$z_1= 4.6*10^{-3}m_1 m ----(1)$

For the second wavelength we have

$z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}$

$z_2 = 0.003825m_2$

$z_2 = 3.825 *10^{-3} m_2 m$ ----(2)

From the question we are told that the two sides coincides with one another so

$zy_1 =z_2$

$4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m$

$\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}$

Hence for this equation to be solved

$m_1 = 3$

and $m_2 = 4$

Substituting this into the equation

$z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}$

Hence $z_1 = z_2 = 0.0138m$

7 0

3 years ago

A student is pushing a box across the room. To push the box three times farther, the student needs to do how much work?

Travka [436]

Answer:

Removing some of the books reduced the mass of the box, and less force was needed to push it across the floor.

8 0

3 years ago