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Zina [86]
2 years ago
9

14.

Physics
1 answer:
larisa [96]2 years ago
7 0

Answer:

The answer is C, air resistance!

Explanation:

A frictional force in the air is air resistance because it is in the air.

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Find the current flowing out of the battery.​
klemol [59]

Answer:

0.36 A.

Explanation:

We'll begin by calculating the equivalent resistance between 35 Ω and 20 Ω resistor. This is illustrated below:

Resistor 1 (R₁) = 35 Ω

Resistor 2 (R₂) = 20 Ω

Equivalent Resistance (Rₑq) =?

Since, the two resistors are in parallel connections, their equivalence can be obtained as follow:

Rₑq = (R₁ × R₂) / (R₁ + R₂)

Rₑq = (35 × 20) / (35 + 20)

Rₑq = 700 / 55

Rₑq = 12.73 Ω

Next, we shall determine the total resistance in the circuit. This can be obtained as follow:

Equivalent resistance between 35 Ω and 20 Ω (Rₑq) = 12.73 Ω

Resistor 3 (R₃) = 15 Ω

Total resistance (R) in the circuit =?

R = Rₑq + R₃ (they are in series connection)

R = 12.73 + 15

R = 27.73 Ω

Finally, we shall determine the current. This can be obtained as follow:

Total resistance (R) = 27.73 Ω

Voltage (V) = 10 V

Current (I) =?

V = IR

10 = I × 27.73

Divide both side by 27.73

I = 10 / 27.73

I = 0.36 A

Therefore, the current is 0.36 A.

6 0
3 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
2 years ago
As a space shuttle climbs, _______
n200080 [17]
Rocket fuel will and smoke will emit from the thrusters.
4 0
3 years ago
Read 2 more answers
What is the formula to find initial velocity?​
Inga [223]

Answer:

s = ut + 1/2at^2

Explanation:

s= distance

u= initial velocity

a= acceleration

7 0
2 years ago
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