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3 years ago

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What does this circuit do? T is the clock All flip flops are D-Flip-Flop What is the function of the circuit if Dser is ‘1’ alwa

ys and flip flops are initialized to ‘001’. What if the initial values are ‘000’ and Dser is ‘0’? What if the initial values are ‘111’ and Dser is ‘1’? Draw T and Qser for each scenario.

1 answer:

Harlamova29_29 [7]3 years ago

6 0

Answer:

That is true

Explanation:

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What are the messing forces that would make the object be in equilibrium?

antoniya [11.8K]

Answer:

A) 20 N, B) 20 N, & C) 8 N

Explanation:

For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.

1. Determination of A and B.

Forward forces = Backward forces

A + 10 + B = 25 + 25

A + 10 + B = 50

Collect like terms

A + B = 50 – 10

A + B = 40

Assume A and B to be equal. Thus, A is 20 N and B is 20 N.

2. Determination of C

Upward forces = Downward forces

C + 112 = 20 + 100

C + 112 = 120

Collect like terms

C = 120 – 112

C = 8 N

Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.

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3 years ago

What energy transformations occur in a tv remote?

Tatiana [17]

Pressing the button: Mechanical to electrical
<span>Waste energy from this step: Mechanical to heat </span>
<span>Converting electrical signal to IR: electrical to electromagnetic (infrared rays) </span>
<span>Waste energy from this step: electrical to heat </span>
<span>Chemical energy from battery: chemical to electrical</span>

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3 years ago

What two types of scientific knowledge can be expressed as mathematical equations? (Please don't tell me the answer could you pl

gtnhenbr [62]

An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.

5 0

4 years ago

You are holding a shopping basket at the grocery store with two 0.60 kg cartons of cereal at one end of the basket. The basket i

ludmilkaskok [199]

Answer:

The gallon of milk is placed at a distance of 0.641 meters.

Explanation:

We have,

Mass of two 0.6 kg cartons of cereal is 1.2 kg

Length of the basket, l = 0.77 m

Center of the basket is 0.385 m

Mass of the half gallon of milk is 1.8 kg

Total weight in the basket is 1.2 + 1.8 = 3 kg

We need to find the position of the gallon of milk so that the center of mass (COM) of your groceries is at the center of the basket.

So,

(3 × 0.385) = (1.2 × 0) + (1.8 × x)

1.155 = 1.8 x

x = 0.641 meters

So, the gallon of milk is placed at a distance of 0.641 meters.

5 0

4 years ago

Drops of rain fall perpendicular to the roof of a parked car during a rainstorm. The drops strike the roof with a speed of 15 m/

IrinaK [193]

Answer: (a) 1.065 N (b) 2.13 N

Explanation:

<h2>(a) average force exerted by the rain on the roof</h2><h2 />

According Newton's 2nd Law of Motion the force $F$ is defined as <u>the variation of linear momentum</u> $p$ <u>in time:</u>

$F=\frac{dp}{dt}$ (1)

Where the linear momentum is:

$p=mV$ (2) Being $m$ the mass and $V$ the velocity.

In the case of the rain drops, which initial velocity is $V_{i}=15m/s$ and final velocity is $V_{f}=0$ (we are told the drops come to rest after striking the roof). The momentum of the drops $p_{drops}$ is:

$p_{drops}=mV_{i}+mV_{f}$ (3)

If $V_{f}=0$ , then:

$p_{drops}=mV_{i}$ (4)

Now the force $F_{drops}$ exerted by the drops is:

$F_{drops}=\frac{dp_{drops}}{dt}=\frac{d}{dt}mV_{i}$ (5)

$F_{drops}=\frac{dm}{dt}V_{i}+m\frac{dV_{i}}{dt}$ (6)

At this point we know the mass of rain per second (mass rate) $\frac{dm}{dt}=0.071 kg/s$ and we also know the initial velocity does not change with time, because that is the velocity at that exact moment (instantaneous velocity). Therefore is a constant, and the derivation of a constant is zero.

This means (6) must be rewritten as:

$F_{drops}=\frac{dm}{dt}V_{i}$ (7)

$F_{drops}=(0.071 kg/s)(15m/s)$ (8)

$F_{drops}=1.065kg.m/s^{2}=1.065N$ (9) This is the force exerted by the rain drops on the roof of the car.

<h2>(b) average force exerted by hailstones on the roof </h2><h2 />

Now let's assume that instead of rain drops, hailstones fall on the roof of the car, and let's also assume these hailstones bounce back up off after striking the roof (this means they do not come to rest as the rain drops).

In addition, we know the hailstones fall with the same velocity as the rain drops and have the same mass rate.

So, in this case the linear momentum $p_{hailstones}$ is:

$p_{hailstones}=mV_{i}+mV_{f}$ (9) Being $V_{i}=V_{f}$

$p_{hailstones}=mV+mV=2mV$ (10)

Deriving with respect to time to find the force $F_{hailstones}$ exerted by the hailstones:

$F_{hailstones}=\frac{d}{dt}p_{hailstones}=\frac{d}{dt}(2mV)$ (10)

$F_{hailstones}=2\frac{d}{dt}(mV)=2(\frac{dm}{dt}V+m\frac{dV}{dt})$ (11)

Assuming $\frac{dV}{dt}=0$ :

$F_{hailstones}=2(\frac{dm}{dt}V)$ (12)

$F_{hailstones}=2(0.071 kg/s)(15m/s)$ (13)

Finally:

$F_{hailstones}=2.13kg.m/s^{2}=2.13N$ (14) This is the force exerted by the hailstones

Comparing (9) and (14) we can conclude the force exerted by the hailstones is two times greater than the force exerted by the raindrops.

5 0

3 years ago