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2 years ago

What is a kind of useful car?

Engineering

2 answers:

tatyana61 [14]2 years ago

7 0

I would recommend a truck. It can haul heavy things.

Nutka1998 [239]2 years ago

3 0

Yes what he/she said a truck is a very useful car if you want to go to work with it many use it for that, it can vary many things:)

You might be interested in

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

2 years ago

A Transmission Control Protocol (TCP) connection is in working order and both sides can send each other data. What is the TCP so

kkurt [141]

Answer:

ESTABLISHED

Explanation:

What is TCP?

A Transmission Control Protocol (TCP) is a communication protocol which allows the exchange of data between computers in a network.

When a Transmission Control Protocol connection is up and running meaning that both sides can send and receive data then the corresponding TCP socket states is known as "ESTABLISHED".

The most common socket states are:

LISTEN:

Before a TCP connection is made, there needs to be a server with a listener that will listen on incoming connection request.

ESTABLISHED:

When a TCP connection is up and running meaning that both sides can send and receive data.

CLOSED:

The CLOSED state means that there is no TCP connection.

There are a total of 11 TCP socket states:

1. LISTEN

2. SYN-SENT

3. SYN-RECEIVED

4. ESTABLISHED

5. FIN-WAIT-1

6. FIN-WAIT-2

7. CLOSE-WAIT

8. CLOSING

9. LAST-ACK

10. TIME-WAIT

11. CLOSED

3 0

3 years ago

If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the cond

Dennis_Churaev [7]

Answer:

The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²

Explanation:

The electric field intensity due to a thin conducting sheet is given by the following formula:

Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)

From this formula:

Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)

We have the following data:

Electric Field Intensity = 1.5 N/C

Permittivity of free space = 8.85 x 10^-12 C²/N.m²

Therefore,

Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)

<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>

Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².

7 0

3 years ago

Una empresa realizó en el ejercicio de compras al contado por valor

Tanzania [10]

Answer:

englishhhh pleasee

Explanation:

we dont understand sorry....

8 0

3 years ago

g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,

irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0

2 years ago