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2 years ago

8

What happens if you leave your car on while pumping gas

1 answer:

Natali5045456 [20]2 years ago

5 0

Answer:

the air you pumped would be off

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A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit

Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

$W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf$

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0

3 years ago

A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta

strojnjashka [21]

Answer:

$\dot W_{out} = 399.47\,kW$

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0$

The work done by the turbine is:

$\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}$

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

$P = 10\,MPa$

$T = 520\,^{\textdegree}C$

$h = 3425.9\,\frac{kJ}{kg}$

Outlet (Superheated Steam)

$P = 1\,MPa$

$T = 280\,^{\textdegree}C$

$h = 3008.2\,\frac{kJ}{kg}$

The work output is:

$\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW$

$\dot W_{out} = 399.47\,kW$

5 0

3 years ago

Under which of the following conditions is a Type B-1 Fire extinguisher required onboard a motorized vessel?

swat32

Answer:

The correct option is;

D. The vessel has closed living spaces onboard

Explanation:

Type B-1 Fire extinguishers

A fire extinguisher is required by the law to be installed in a boat that hs the following specifications

1) There are closed compartment in the boat that can be used for fuel storage

2) There exist double double bottom that is only partially filled with flotation materials

3) There are closed living spaces in the boat

4) The fuel tank is permanently installed in the boat

5) The engine is inboard.

7 0

4 years ago

Read 2 more answers

The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit

zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

$T = \pi \sqrt{\frac{l}{g} }$

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

$T = M^{a}l^{b}g^{c}$

$T = KM^{a}l^{b}g^{c}$

where k is the constant of proportionality.

Now putting the dimensional formula:

$T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}$

$M^{0}L^{0}T^{1} = KM^{a}L^{b+c}$

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

$T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }$

T = $2\pi \sqrt{\frac{l}{g} }$

where k = $2\pi$

8 0

3 years ago

1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn

emmasim [6.3K]

Answer:

1. $\dot Q=19600\ W$

2. $\dot Q=120\ W$

Explanation:

1.

Given:

height of the window pane, $h=2\ m$

width of the window pane, $w=1\ m$

thickness of the pane, $t=5\ mm= 0.005\ m$

thermal conductivity of the glass pane, $k_g=1.4\ W.m^{-1}.K^{-1}$

temperature of the inner surface, $T_i=15^{\circ}C$

temperature of the outer surface, $T_o=-20^{\circ}C$

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

$\dot Q=k_g.A.\frac{dT}{dx}$

here:

A = area through which the heat transfer occurs = $2\times 1=2\ m^2$

dT = temperature difference across the thickness of the surface = $35^{\circ}C$

dx = t = thickness normal to the surface = $0.005\ m$

$\dot Q=1.4\times 2\times \frac{35}{0.005}$

$\dot Q=19600\ W$

2.

air spacing between two glass panes, $dx=0.01\ m$

area of each glass pane, $A=2\times 1=2\ m^2$

thermal conductivity of air, $k_a=0.024\ W.m^{-1}.K^{-1}$

temperature difference between the surfaces, $dT=25^{\circ}C$

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

$\dot Q=k_a.A.\frac{dT}{dx}$

$\dot Q=0.024\times 2\times \frac{25}{0.01}$

$\dot Q=120\ W$

5 0

3 years ago