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3 years ago

An integrated circuit is a _____.

Physics

2 answers:

Mademuasel [1]3 years ago

8 0

closed and unsecure circuit

MatroZZZ [7]3 years ago

7 0

An integrated circuit is a set of electronic circuits on one small flat piece of silicon

i hope this help. good luck

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A ceiling fan draws a current of 0.625 A and has a voltage of 120 V. The resistance of the ceiling fan, to the nearest whole num

Fiesta28 [93]

So we want to know what is the resistance of the ceiling fan if the current I=0.625 A and the voltage V=120 V. So we can get the resistance from Ohms law: I=V/R where I is the current, V is voltage and R is resistance. So the resistance is R=V/I. Now we input the numbers and get: R=120/0.625=192 Ω. That is resistance to the closest whole number.

5 0

3 years ago

Read 2 more answers

A cycle travels along a circular track of diameter 42 m. Calculate the distance travelled and the displacement of the cycle in (

DENIUS [597]

Answer:

(a) i) The distance travelled by the cycle in half round is approximately 65.97 m

ii) The displacement is 42 m

(b) (i) The distance travelled in one round is approximately 131.95 m

(ii) The displacement of the cycle in one round is 0

Explanation:

The diameter of the track through which the cycle travels, D = 42 m

(a) i) Half round is the motion of half the length of the circular path

The distance travelled by the cycle in half round = The length of half the circular track = (1/2) × π × D

∴ The distance travelled by the cycle in half round = (1/2) × π × 42 m = 21·π m ≈ 65.97 m

ii) The displacement half round = The change in the location of the cycle = The difference between the start and stop locations of the cycle on a straight line after half round

The angle at the center of the circular path the cycle turns in half round = 180°

Therefore, the path between the start and stop location of the cycle in half round = The diameter of the circular track

The displacement of the cycle in half round = The diameter of the circular track = 42 m

(b) (i) The distance travelled in one round = The perimeter of the circular track = π·D

∴ The distance travelled in one round = π × 42 m ≈ 131.95 m

(ii) The displacement of the cycle in one round = The change in the location of the cycle

The start and stop location of the cycle after moving one round is the same, therefore, there is no change in the location of the cycle.

Therefore we have;

The displacement of the cycle in one round = 0 (no change in location of the cycle)

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3 years ago

The number of times that a wave vibrates in unit is called

kari74 [83]

I believe it's frequency.

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4 years ago

The Earth moving round the Sun in a circular orbit is acted upon by a

zalisa [80]

No I do not agree. It is because work is done when force acting o a body displace or covers certain displacement in the direction of force applied .

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3 years ago

In a women's 100-m race, accelerating uniformly, Laura takes 1.82 s and Healan 3.07 s to attain

Sav [38]

Answer:

Laura is ahead and for a distance of 3.22 m

Explanation:

To solve this problem of one-dimensional kinematics, we have to find the acceleration and the final speed of each runner. Let's start with Laura

Lura data is acceleration time 1.82s, total run time 10.4 s and total distance 100m. In all the races the rest starts, so the initial speed is zero (Vo = 0)

Vf1= Vo + a1 t1

Vf1 = x/t

XT = X1 + X2

X1 = Vo t1 + ½ a1 t1²

X1 = ½ a1 t1²

X2 = Vf1 (t-t1)

This is the remaining time of the race after the acceleration is over.

XT = ½ a1 t1² + Vf1 (t-t1)

We remplace the expression of Vf1

XT = ½ a1 t1² + a1 t1 (t-t1)

Laura's aceleration (a1) is

a1= XT / [ ½ t1² + t1 (t-t1)]

a1= 100/ [ ½ 1.82²+ 1.82 (10.4 -1.82)]

a1= 5.79m/s2

We repeat the same calculation for the other Healan runner, whose data are: total distance 100m, acceleration time 3.07 s and total time 10.4 s

Vf2= Vo + a2 t2

Vf2 = x/t

XT = X3 + X4

X3 = Vo t2 + ½ a2 t2²

X3 = ½ a2 t2²

X4 = Vf2 (t-t2)

XT = ½ a2 t2² + Vf2 (t- t2)

XT = ½ a2 t2² + a2 t2 (t-t2)

The aceleration of Healan (a2)

a2 = XT / [½ t2² + t2 (t-t2)]

a2 = 100 / [½ 3,07²+ 3.07 (10.4 -3.07)]

a2 = 3.67 m / s2

We also need the final speeds of each runner

Laura Vf1 = Vo + a1 t1

Vf1 = 0 + 5.79 1.82

Vf1 = 10.54 m / s

Healan Vf2 = Vo + a2 t2

Vf2 = 0 + 3.67 3.07

Vf2 = 11.27 m / s

Having the acceleration and speed of each runner, you can start answering the questions

a) For t3 = 6.15s

Laura

The time to stop with constant speed is what remains after accelerating

XL= ½ a1 t1² + Vf1 (t3-t1)

XL= ½ 5.79 1.82² + 10.54 (6.15 – 1.82)

XL= 55.23 m

Healan

XH= ½ a2 t2² + Vf2 (t3-t2)

XH= ½ 3.67 3.07² + 11.27 (6.15-3.07)

XH= 52.01 m

(XL -XH)= 55.23- 52.01

(XH -XL)= 3.22 m

It is appreciated from these results that Laura is ahead and for a distance of 3.22 m

b) If we analyze the acceleration values of each runner, knowing that they leave the rest and that Healan at the end has a speed greater than Laura, the point of maximum distance difference is when Laura stops accelerating t = 1.82 s

XL= ½ a1 t12

XL= ½ 5.79 1.822

XL= 9.59 m

XH = ½ a2 t12

XH= ½ 3.67 1.822

XH= 6.08 m

The maximum distance difference is 3.51 m

c) Already analyzed in the previous part 1.82 s, since the Laura stop accelerating and Heala continue with acceleration will travel greater distances in equal time units

6 0

4 years ago