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Wewaii [24]
3 years ago
13

Convert 25 mm into in.

Engineering
1 answer:
mihalych1998 [28]3 years ago
4 0

Answer:

25 mm = 0.984252 inches

Explanation:

Millimeter and inches are both units of distance. The conversion of millimeter into inches is shown below:

<u>1 mm = 1/25.4 inches</u>

From the question, we have to convert 25 mm into inches

Thus,

<u>25 mm = (1/25.4)*25 inches</u>

So,

25 mm=\frac{25}{25.4} inches

Thus, solving we get:

<u>25 mm = 0.984252 inches</u>

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When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs
Reika [66]

Answer:\mu_{k}

Explanation:

We use kinetic friction when a body is moving i.e. \mu_{k} for calculations.

Static friction is used when a body is in rest while kinetic friction is used when a body is moving and its value is quite low as compared to static friction .

Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

3 0
3 years ago
You are traveling upstream on a river at dusk. You see a buoy with the number 5 and a flashing green light . What should you do?
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Explanation:

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3 years ago
What is the force in kN of work done is 1.2 ms moves through 120m​
Semmy [17]

Answer:

\frac{1.2}{120}

0.01

5 0
2 years ago
Interpret the assembly program below: MOV R3,R0;
Reika [66]

Answer:

Explanation:

1.  With the operands R0, R1, the program would compute AND operation and ADD operation .

2. The operands could truly be signed 2's complement encoded (i.e Yes) .

3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.

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7 0
3 years ago
The convection coefficient for flow over a solid sphere may be determined by submerging the sphere, which is initially at 25 °C,
sashaice [31]

Answer:

t = 59.37 s

Explanation:

Given data:

thermal diffusivity = \alpha = \frac{k}{\rho c_p} =0.40\times 10^{-0.5}

theraml conductivity = k = 22 W/m.K

h = 300 W/ m^2.K

T_i = 25 degree C = 298 k

T_o = 60 degree C = 333 k

T_{\infty}= 75 degree C =  348 L

diameter d = 0.1 m

characteristics length Lc = r/3 = = 0.0166

Bi = \frac{hLc}{K} = \frac{300\times 0.0166}{22} = 0.226

\tau = \frac{\alpha t}{lc^2} = \frac{0.4\times 10^{-5}\times t}{0.0166^2}

\tau = 0.036 t

\frac{T_o -T_{\infty}}{T_i -T_{\infty}} = Ae^[\lambda^2 \tau}

at Bi = 0.226

Ai = 0.982

\lambda = 0.876

\frac{333348}{298-348} = 0.982e^{-0.879^2 0.036t}

0.3 = 0.982 e^{-0.2t}

0.305 = e^{-0.2t}

-1.187 = - 0.02t

t = 59.37 s

7 0
3 years ago
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