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Nonamiya [84]
2 years ago
12

un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente

el sol habia elevado la temperatura habia elevado los 131 fareinheit ¿cuanta gasolina derramo el tanque?
Physics
1 answer:
Sedaia [141]2 years ago
3 0

Answer:

Volume of gasoline that expands and spills out is 1.33 ltr

Explanation:

As we know that when temperature of the liquid is increased then its volume will expand and it is given as

\Delta V = V_o\gamma \Delta T

here we know that

V_o = 40 Ltr

volume expansion coefficient of the gasoline is given as

\gamma = 950 × 10^{–6}

change in temperature is given as

\Delta T = (131 - 68) \times \frac{5}{9}

\Delta T = 35 ^oC

Now we have

\Delta V = 40(950 \times 10^{-6})(35)

\Delta V = 1.33 Ltr

You might be interested in
the steel bed of a suspension bridge is 200m long at 20 C. If the extremes of temperature to which it might be exposed are -30 C
raketka [301]

Answer:

The steel bed will contract by  0.13 m, and expand by 0.052 m

Explanation:

For contraction,

α = ΔL/(LΔθ)..................... Equation 1

Where α = Linear expansivity of  steel, ΔL = decrease in length/ Increase in length, L = Original length, Δθ = Change in temperature

make ΔL the subject of the equation

ΔL = α(LΔθ)................. Equation 2

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = -30-20 = -50 °C

Substitute into equation 2

ΔL = 13×10⁻⁶(200)(-50)

ΔL = -0.13 m

Similarly, For expansion,

Using equation 2

ΔL =  α(LΔθ)

Given: α = 13×10⁻⁶/C, L = 200 m, Δθ = 40-20 = 20  °C

Substitute into equation 2

ΔL =  13×10⁻⁶(200)(20)

ΔL = 0.052 m.

Hence the steel bed will contract by  0.13 m, and expand by 0.052 m

3 0
2 years ago
89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a
ddd [48]

Explanation:

Given that,

Initial speed of the electron, u=4\times 10^5\ m/s

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, a=6\times 10^{12}\ m/s^2  

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

v^2=u^2+2as

v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05

v = 871779.788 m/s

or

v=8.71\times 10^5\ m/s

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

t=\dfrac{v-u}{a}

t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}

t=7.85\times 10^{-8}\ s

Hence, this is the required solution.

4 0
2 years ago
B. A hydraulic jack has a ram of 20 cm diameter and a plunger of 3 cm diameter. It is used for lifting a weight of 3 tons. Find
lozanna [386]

Answer:

option (b)

Explanation:

According to the Pascal's law

F / A = f / a

Where, F is the force on ram, A be the area of ram, f be the force on plunger and a be the area of plunger.

Diameter of ram, D = 20 cm, R = 20 / 2 = 10 cm

A = π R^2 = π x 100 cm^2

F = 3 tons = 3000 kgf

diameter of plunger, d = 3 cm, r = 1.5 cm

a = π x 2.25 cm^2

Use Pascal's law

3000 / π x 100 = f / π x 2.25

f = 67.5 Kgf

4 0
2 years ago
1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.
Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.  She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h

Hence, this is the required solution.

8 0
2 years ago
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0
2 years ago
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