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2 years ago

What reading will a digital multimeter indicate, when set to the 4A scale, if the actual current flow is 200mA?0.002 A

Engineering

1 answer:

Andreyy892 years ago

6 0

Answer:

Current flow in ampere according to 4A scale = 0.2 Ampere

Explanation:

Given;

Current flow = 200 milliampere

Find:

Current flow in ampere according to 4A scale

Computation:

We know that

1 ampere = 1,000 milliampere

So,

1 milliampere = 1/1,000 ampere

Current flow = 200 milliampere

Current flow = 200[1/1,000]

Current flow = 200[0.001]

Current flow = 0.2

Current flow in ampere according to 4A scale = 0.2 Ampere

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The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

7 0

2 years ago

Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made

Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random.Therefore, number ways in which cars can line up is given as:

No. of ways = 20! = 2.43 x 10^18

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

Probability = 0.00001 = 0.001%

4 0

2 years ago

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 18 kN are appli

monitta

The smallest allowable depth is $d=16.04 \mathrm{mm}$ for the milled portion of bar.

Explanation:

Given,

Magnitude of force, $\mathbf{p}=18 \mathrm{kN}$

$a=30 \mathrm{mm}$

$=0.03 \mathrm{m}$

Allowable stress, $\sigma_{a l l}=135 \mathrm{MPa}$

cross sectional area of bar,

$A=a \times d$

$A=a d$

e - eccentricity

$e=\frac{a}{2}-\frac{d}{2}$

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$M=P e$

$=P\left(\frac{a}{2}-\frac{d}{2}\right)$

$=\frac{P(a-d)}{2}$

Allowable stress

$\sigma=\frac{P}{A}+\frac{M c}{I}$

$c=\frac{d}{2}$

Moment of Inertia,

$I=\frac{b d^{3}}{12}$

$=\frac{a d^{3}}{12}$

$\therefore \sigma=\frac{P}{a d}+\frac{\frac{P(a-d)}{2} \times \frac{d}{2}}{\frac{a d^{3}}{12}}$

$\sigma=\frac{P}{a d}+\frac{3 P(a-d)}{a d^{2}}\\$

$\sqrt{x} \sigma\left(a d^{2}\right)=P d+3 P(a-d)$

$\sigma\left(a d^{2}\right)=P d+3 P a-3 P d$

$\sigma\left(a d^{2}\right)=(P-3 P) d+3 P a$

$\left(\sigma a d^{2}\right)=-2 P d+3 P a$

$\sigma d^{2}=-\frac{2 P}{a} d+3 P$

By substituting values we get,

$\left(135 \times 10^{6}\right) d^{2}+\frac{2 \times 18 \times 10^{3}}{0.03} d-3\left(18 \times 10^{3}\right)=0$

$\left(135 \times 10^{6}\right) d^{2}+\left(12 \times 10^{5}\right) d-54 \times 10^{3}=0$

On solving above equation we get, $d=0.01604 \mathrm{m}\\$

$d=16.04 \mathrm{mm}$

3 0

2 years ago

A spherical hot-air balloon is initially filled with air at 120 kPa and 20°C with an initial diameter of 5 m. Air enters this ba

adoni [48]

Answer:

time is 17.43 min

Explanation:

given data

initial diameter = 5 m

velocity = 3 m/s

final diameter = 17 m

solution

we will apply here change in change in volume equation that is express as

ΔV = $\frac{4}{3} \pi * (rf)^3 - \frac{4}{3} \pi * (ri)^3$ .............1

here ΔV is change in volume and rf is final radius and ri is initial radius

ΔV = $\frac{4}{3} \pi * (8.5)^3 - \frac{4}{3} \pi * (2.5)^3$

ΔV = 2507 m³

Q = velocity × Area

Q = 3 × π ×(0.5)² = 2.356 m³/s

and

change in time is express as

Δt = $\frac{\Delta V}{Q}$

Δt = $\frac{2507}{2.356}$

Δt = 1046 sec

so change in time is 17.43 min

8 0

2 years ago

2. What is the charge, expressed in micro coulombs on two equally and similarly charges spheres placed in air with their centres

9966 [12]

Answer:

q = 0.1086 micro Coulombs

Explanation:

By Coulombs law, we have;

$F =k \times \dfrac{ q_1 \times q_2}{r^2}$

Where;

F = The electric force = 120 mgm

q₁ and q₂ = Charge

r = The separating distance = 30 cm = 0.3 m

k = 8.9876×10⁹ kg·m³/(s²·C²)

Where, q₁ and q₂, we have;

$F =k \times \dfrac{ q^2}{r^2}$

Whereby the force is the force of 120 milligram mass, we have;

0.00012 × 9.81 = 000011772 N

$q = \sqrt{ \dfrac{ F\times r^2}{k}}$

Substituting the values, we have;

$q = \sqrt{ \dfrac{ 000011772 \times (0.3)^2}{8.9876 \times 10^9}} = 1.086 \times 10 ^{-7} \ Coulombs$

q = 0.1086 micro Coulombs.

5 0

2 years ago