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aniked [119]
1 year ago
15

The fuel for a jet engine has a density of 1.32 slug/ft3. if the total volume of fuel tank a is 50 ft3, determine the weight of

the fuel when the tanks are completely full?
Physics
1 answer:
inessss [21]1 year ago
6 0

The weight of the fuel when the tank is filled is 2.13 kip.

The density of the fuel of the jet engine= 1.32 slug/ft³

The total volume of the fuel that can be filled in the fuel tank of a jet engine = 50 ft³

The specific weight of the fuel when the tank is filled is,

\gamma  =  p \times g

\gamma  = 1.32 \times   32.2

\gamma  = 42.504 \: ib/ft ^{3}

The specific weight of the fuel when the tank is filled is 42.504 ib/ft ³.

The weight of the fuel when the tank is filled is,

w =  \gamma  \times ∀

w  = 42.5.4 \times 50

w = 2.13 \: kip

The weight of the fuel is 2.13 kip.

Therefore, the weight of the fuel when the tank is filled is 2.13 kip.

To know more about specific weight, refer to the below link:

brainly.com/question/13178675

#SPJ4

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' A ' and ' D ' are both correct statements.

3 0
3 years ago
If an atom has an atomic number of 2, it will be stable with 2 electrons in its valence shell. Group of answer choices True Fals
torisob [31]

Answer: True

Explanation:

Atomic number is defined as the number of protons or the number of electrons that are present in an electrically neutral atom.

Atomic number = Number of protons = number of electrons = 2

Electronic configuration represents the total number of electrons that a neutral element contains. We add all the superscripts to know the number of electrons in an atom.

The electronic configuration will be 1s^2

As its duplet is already complete and it has noble gas configuration , it is stable with 2 valence electrons.

3 0
3 years ago
You serve a volleyball with a mass of 2.5 kg. The ball leaves your hand with a speed of 23 m/s. What is the kinetic energy of th
Zarrin [17]

Answer:

661.25, I believe!

Explanation:

KE = (1/2)(2.5)(23)^2

KE=(1.25)(529)

KE= 661.25

8 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
⦁ A car going 50 m/s is brought to rest in a distance of 20.0 m as it strikes a pile of dirt. How large an average force is exer
gtnhenbr [62]

Answer:

the average force exerted by seatbelts on the passenger is 5625 N.

Explanation:

Given;

initial velocity of the car, u = 50 m/s

distance traveled by the car, s = 20 m

final velocity of the after coming to rest, v = 0

mass of the passenger, m = 90 kg

Determine the acceleration of the car as it hit the pile of dirt;

v² = u² + 2as

0 = 50² + (2 x 20)a

0 = 2500 + 40a

40a = -2500

a = -2500/40

a = -62.5 m/s²

The deceleration of the car is 62.5 m/s²

The force exerted on the passenger by the backward action of the car is calculated as follows;

F = ma

F = 90 x 62.5

F = 5625 N

Therefore, the average force exerted by seatbelts on the passenger is 5625 N.

8 0
3 years ago
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