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2 years ago

Which of the following statements is TRUE in regards to Signal #2?

Physics

1 answer:

Marina CMI [18]2 years ago

7 0

Answer:

2) Signal #2 is a digital signal that transmits signals in segments/intervals.

Explanation:

Signals are form of information propagating from the source to a display unit for appropriate interpretation. It can be either in a digital or analogue form.

A digital signal is a definite signal which is discrete in time and amplitude. It is mostly in the form of codes obtained from set of values. Graphically it transmits in the form of 1 and 0, showing a point of maximum amplitude (1) and minimum amplitude (0).

Analog signal is continuous signal describing the variation of two variables with respect to time.

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A form of erosion in which particles of sand or dust rub across the surface of rocks.

gayaneshka [121]

The answer is B) Abrasion.

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2 years ago

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if you crash your car how could you decrease the damage to you or the car using the concept of impulse

kotykmax [81]

Explanation:

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2 years ago

A speed-time graph is shown below:

VladimirAG [237]

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)

$\Delta t=8 s$

Then:

$a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}$

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2 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :