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Natalka [10]
3 years ago
6

Convert 250 lb·ft to N.m. Express your answer using three significant figures.

Engineering
1 answer:
vfiekz [6]3 years ago
6 0

Answer:

It will be equivalent to 338.95 N-m

Explanation:

We have to convert 250 lb-ft to N-m

We know that 1 lb = 4.45 N

So foe converting from lb to N we have to multiply with 4.45

So 250 lb = 250×4.45 =125 N

And we know that 1 feet = 0.3048 meter

Now we have to convert 250 lb-ft to N-m

So 250lb-ft=250\times 4.45N\times 0.348M=338.95N-m

So 250 lb-ft = 338.95 N-m

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There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

                                        u_e = 9.30777 * 10^-8  J/m^3

5 0
3 years ago
Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha
Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 0
3 years ago
Design a plate and frame heat exchanger for the following problem:
qwelly [4]

Answer:

See explaination and attachment.

Explanation:

Iteration method is a repetitive method applied until the desired result is achieved.

Let the given equation be f(x) = 0 and the value of x to be determined. By using the Iteration method you can find the roots of the equation. To find the root of the equation first we have to write equation like below

x = pi(x)

Let x=x0 be an initial approximation of the required root α then the first approximation x1 is given by x1 = pi(x0).

Similarly for second, thrid and so on. approximation

x2 = pi(x1)

x3 = pi(x2)

x4 = pi(x3)

xn = pi(xn-1).

please go to attachment for the step by step solution.

8 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
An L2 steel strap having a thickness of 0.125 in. and a width of 2 in. is bent into a circular arc of radius 600 in. Determine t
lesya692 [45]

Answer:

the maximum bending stress in the strap is 3.02 ksi

Explanation:

Given the data in the question;

steel strap thickness = 0.125 in

width = 2 in

circular arc radius = 600 in

we know that, standard value of modulus of elasticity of L2 steel is; E = 29 × 10³ ksi;

Now, using simple theory of bending

1/p = M/EI

solve for M

Mp = EI

M = EI / p ----- let this be equation 1

The maximum bending stress in the strap is;

σ = Mc / I -------let this be equation 2

substitute equation 1 into 2

σ = ( EI / p)c / I

σ = ( c/p )E

so we substitute in our values

σ = ( (0.125/2) / 600 )29 × 10³

σ = 0.00010416666 × 29 × 10³

σ = 3.02 ksi

Therefore, the maximum bending stress in the strap is 3.02 ksi

3 0
2 years ago
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