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3 years ago

Which of the following components of an electrical circuit is generally used to connect all other components?

1 answer:

3 0

A a wire, a wire is connected between the positive and negative ends of the eletrical energy source and connects all the components in between.

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You hang a heavy ball with a mass of 30 kg from a tungsten rod 2.8 m long by 1.5 mm by 2.6 mm. You measure the stretch of the ro

guajiro [1.7K]

Answer:

Young's modulus (Y) = 3.56×10^11 N/m^2

The speed of sound in tungsten = 6166.4 m/s

Explanation:

Young's modulus (Y) = stress/strain

Stress = force/area

Force = mg = 30×9.8 = 294 N

Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2

Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2

Strain = extension/length

Extension = 0.000594 m

Length = 2.8 m

Strain = 0.000594/2.8 = 2.12×10^-4

Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2

Y = h × rho × g

rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3

h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m

From the equations of motion

v^2 = u^2 + 2gh =

Initial speed (u) = 0 m/s

v = sqrt (2×9.8×1.94×10^6)

v = 6166.4 m/s

7 0

3 years ago

Read 2 more answers

Why are all bodies in nature electrically neutral? short answer please

DENIUS [597]

Answer: because they have equal numbers of protons

Explanation: Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion

3 0

2 years ago

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

3 years ago

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operat

omeli [17]

Answer:

The maximum permissible propagation delay per flip flop stage is<u> 100 </u>n sec

Explanation:

1024 ripple counter has 10 J-K flip flops(210 = 1024).

So the total delay will be 10×x where x is the delay of each J-K flip flops.

The period of the clock pulse is 1× 10⁻⁶ s.

Now

10x <= 10⁻⁶ s

x <= 100 ns

x= 100 ns for prpoer operation.

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is <u>100 </u>n sec.

4 0

3 years ago

A force of 20 N acts toward the east. Another force of 27 N acts at the same point toward the west. What is the magnitude and di

Jet001 [13]

Since we are working in one dimension (left right or East West), we don't need to worry about angles! It's just simply a matter of adding things up!

First list out all the forces and add negative (-ive) signs to each of the 'west' forces like this.

20 East + (-27 West) + ? = 10 East

so it's easy to see that 20 + (-27) = -7
So to get to 10 from -7 just do the sum to get 17.
Since 17 is not negative it must be in the direction of East.
So the answer is:
Magnitude = 17 N
Direction = toward the East

8 0

3 years ago