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Zigmanuir [339]
3 years ago
15

A resistance is added in parallel to a 470 Ω resistance to give an effective resistance of 330 Ω. What is the approximate value

of the added resistance? A) 140 Ω B) 680 Ω C) 1.1 kΩ D) 1.8 kΩ
Physics
1 answer:
saw5 [17]3 years ago
4 0
I think its c on the test I had
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dusya [7]
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6 0
3 years ago
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redacteaza un text in care sa iti exprimi parerea in legaturacurtamentul golegilor din textul oracolul
Ivahew [28]

Answer:

Seriously I have no idea. I need help with my homework.

Explanation:

I really need help with my homework. Sorry

4 0
3 years ago
A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while los
mafiozo [28]

Answer:

Option B

Change in entropy of the process is \Delta S= Rln(\frac{P_{1}}{P_{2}})

Explanation:

The entropy of a system is a measure of the degree of disorderliness of the system.

The entropy of a system moving from process 1 to 2 is given as

\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}

recall from first law, \delta q =du +Pdv

hence we have, \Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}

since the process is isothermal, du= 0

this gives us \Delta S = \int\limits^2_1 {\frac{Pdv}{T}}

integrating within the limits of 1 and 2, will give us

\Delta S = R ln (\frac{V_{2}}{V_{1}})

also from ideal gas laws,

\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}

hence we have    \Delta S = R ln (\frac{P_{1}}{P_{2}})

This makes the correct option B

4 0
3 years ago
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2. A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.5 . At 30.0 s after blast
harina [27]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

6 0
3 years ago
It has been suggested that a heat engine could be developed that made use of the fact that the temperature several hundred meter
lesantik [10]

Answer:

efficiency = 5.4%

Explanation:

Efficiency of heat engine is given as

\eta = \frac{W}{Q_{in}}

now we will have

W = Q_1 - Q_2

so we will have

\eta = 1 - \frac{Q_2}{Q_1}

now we know that

\frac{Q_2}{Q_1} = \frac{T_2}{T_1}

so we have

\eta = 1 - \frac{T_2}{T_1}

\eta = 1 - \frac{273+6}{273+22}

\eta = 0.054

so efficiency is 5.4%

8 0
3 years ago
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