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Nataly [62]
3 years ago
15

An airplane is flying horizontally with a constant momentum during a time interval ?t. (a) Is there a net impulse acting on the

plane during this time? Use the impulse-momentum theorem to guide your thinking. (b) In the horizontal direction, both the thrust generated by the engines and air resistance act on the plane. Considering your answer to part (a), how is the impulse of the thrust related in magnitude and direction to the impulse of the force due to the air resistance?
Physics
1 answer:
goldenfox [79]3 years ago
7 0

Answer:

Explanation:

Given

Airplane is flying with horizontally with a constant momentum during time interval \Delta t

Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant

(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction                            

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
nikklg [1K]

Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2} (14)

a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

4 0
3 years ago
An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =
Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V_m = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V_m =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

I(t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

I(t) = V₀√(C/L)

we substitute

I(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

I(t) = 25 × 0.00002

I(t) = 0.0005 A

I(t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0
3 years ago
What are the characteristics of the image based on the values?Check all that apply
Ahat [919]

inverted, real, behind the lens is the answer

8 0
3 years ago
Read 2 more answers
Q:
Shalnov [3]

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

36=\frac{1}{2}m(12)^2 and isolating the m:

m=\frac{2(36)}{12^2} which gives us

m = .50 kg

3 0
3 years ago
A wire is joined to points X and Y in the circuit diagram shown. A diagram of a circuit with a power source on the left. Directl
Sidana [21]

The circuit change when the wire is added will see a short circuit occur and makes bulbs 1 and 2 turn off but keeps bulbs 3 and 4 lit. Option D. This is further explained below.

<h3>How does the circuit change when the wire is added?</h3>

Generally, Electronic circuits consist of a series of interconnected parts that form a closed loop through which electricity may flow.

In conclusion, If two wires are linked together, a short circuit will develop, cutting power to bulbs 1 and 2. But there is no impact on bulbs 3 and 4. There is no problem with bulbs 3 and 4.

Read more about circuit

brainly.com/question/21505732

#SPJ1

7 0
2 years ago
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