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alexandr1967 [171]
3 years ago
8

A rocket is being launched straight up. Air resistance is not negligible.

Physics
2 answers:
devlian [24]3 years ago
7 0

A rocket is launching upwards. The force acting on the rocket to move upward is F. The forces acting opposite to thrust are weight and drag force. The drag force is due to the air resistance. In order to move upwards the thrust force must be greater than weight and drag force. Figure shows the free body diagram of rocket at the time of launching.There are only three forces: weight (always present), thrust (provided from the rocket’s engines to lift it up) and air resistance (said specifically to be not negligible). There are no horizontal forces.

Yanka [14]3 years ago
3 0

Answer:

hope this will help you

have a great day

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Ivahew [28]

Answer:

0.147 J

Explanation:

The total energy that has been transformed into thermal energy is equal to the loss of gravitational potential energy between the initial situation (bob at h=0.5 m above the ground) and the final situation (bob back but at h=0.45 m above the ground).

Therefore, we have

E_{thermal}=\Delta U=mgh_1 - mgh_2 = mg(h_1 -h_2)

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g = 9.8 m/s^2

h1 = 0.5 m is the initial height

h2 = 0.45 m is the final height

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E_{thermal}=(0.3 kg)(9.8 m/s^2)(0.5 m-0.45 m)=0.147 J

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How does the tide cycle affect erosion along a sea coast?
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An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth
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Note: The answer choices are :

a) Increased

b) Decreased

c) stayed the same

Answer:

The correct option is Increased

The magnitude of the electric field potential difference between the wingtips increases.

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where l = length

v = velocity

B = magnetic field

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A parallel-plate, air-gap capacitor has a capacitance of 0.14 mu F. The plates are 0.5 mm apart, What is the area of each plate?
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Answer:

7.9060 m²

8.57 Volts

5.142×10⁻⁶ Joule

1.2×10⁻⁶ Coulomb

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C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F

d = Distance between plates = 0.5 mm = 0.5×10⁻³ m

Q = Charge = 1.2 μC = 1.2×10⁻⁶ C

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Capacitance

C=\frac{\epsilon_{0}A}{d}\\\Rightarrow A=\frac{Cd}{\epsilon_{0}}\\\Rightarrow A=\frac{0.14\times 10^{-6}\times 0.5\times 10^{-3}}{8.854\times 10^{-12}}\\\Rightarrow A=7.9060\ m^2

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Voltage

V=\frac{Q}{C}\\\Rightarrow V=\frac{1.2\times 10^{-6}}{0.14\times 10^{-6}}\\\Rightarrow V=8.57\ Volts

∴ Potential difference between the plates if the capacitor is charged to 1.2 μC  is 8.57 Volts.

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∴ Stored energy is 5.142×10⁻⁶ Joule

Charge

Q = CV

⇒Q = 0.14×10⁻⁶×8.57

⇒Q = 1.2×10⁻⁶ C

∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb

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