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3241004551 [841]
3 years ago
12

If you weighed 100 lb on Earth, what would you weigh at the upper atmosphere of Jupiter? For reference, Jupiter has a mass that

is about 300 times Earth’s mass and a radius that is 10 times Earth’s radius.
Physics
1 answer:
nignag [31]3 years ago
7 0

Answer:

The answer to the question is

A 100 lb person would weigh 300.33 lbf at the upper atmosphere of Jupiter

Explanation:

To solve the question we note that

Mass of object = 100 lb =‪ 45.35924‬ kg

Mass of Jupiter = 300×Mass of Earth = 300×5.972 × 10²⁴ kg =1.7916×10²⁷ kg

Radius of Jupiter = 10× Radius of Earth = 10×6,371 km = 63710 km

Gravitational constant, G = 6.67408 × 10⁻¹¹ m³ kg-1 s-2

Gravitational force is given by F_G= \frac{Gm_1m_2}{r^2}

Plugging in the values we get

F_G = \frac{6.67408*10^{-11}*45.35924*1.7916*10^{27}}{63710^2} = 1335.93 N

Converting into lbf gives 1335.93 N *0.2248 lbf/N = 300.33 lbf

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A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If
Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R =  60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀  = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

I(t) = I_o(1-e^{-\frac{t}{T}})

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A

Therefore, the current in the circuit 7 ms later is 0.2499 A

6 0
3 years ago
A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 3.46 m long, weighs 12.8 N, and ro
Mkey [24]

Answer:

a. Rotational inertia: 5.21kgm²

b. Magnitude of it's angular momentum: 123.32kgm²/s

Explanation:

Length of the rod = 3.46m

Weight of the rod = 12.8 N

Angular velocity of the rod= 226 rev/min

a. Rotational Inertia (I) about its axis

The formula for rotational inertia =

I = (1/12×m×L²) + m × ( L ÷ 2)²

Where L = length of the rod

m = mass of the rod

Mass of the rod is calculated by dividing the weight of the rod with the acceleration due to gravity.

Acceleration due to gravity = 9.81m/s²

Mass of the rod = 12.8N/ 9.81m/s²

Mass of the rod = 1.305kg

Rotational Inertia =

(1/12× 1.305 × 3.46²)+ 1.305 ( 3.46÷2)²

Rotational Inertia =  1.3019115 + 3.9057345

Rotational Inertia = 5.207646kgm²

Approximately = 5.21kgm²

b. The magnitude of the rod's angular momentum about the rotational axis is calculated as

Rotational Inertia about its axis × angular speed of the rod.

Angular speed of the rod is calculated as= (Angular velocity of the rod × 2π)/60

= (226×2π) /60

= 23.67 rad/s

Rotational Inertia = 5.21kgm²

The magnitude of the rod's angular momentum about the rotational axis

= 5.21kgm²× 23.67 rad/s

= 123.3207kgm²/s

Approximately = 123.32kgm²/s

7 0
3 years ago
At a distance of 8 m, the sound intensity of one speaker is 66 dB. If we were to place 3 speakers in a circle of radius 8 m, wha
ludmilkaskok [199]

Answer:

dβ = 70. 77 dβ

Explanation:

The intensity of sound in decibels is

         dβ = 10 log I/I₀

let's look for the intensity of this signal

         I / I₀ = 10 dβ/10

         I / I₀ = 3.981 10⁶

the threshold intensity of sound for humans is I₀ = 1 10⁻¹² W / m²

         I = 3.981 10 ⁶ 1 10⁻¹²

         I = 3,981 10⁻⁶ W / m²

It is indicated that 3 cornets are placed in the circle, for which total intensity is

        I_total - 3 I

        I_total = 3  3,981 10⁻⁶

        I_total = 11,943 10⁻⁶ W / m²

let's reduce to decibels

      dβ = 10 log (11,943 10⁻⁶/1 10⁻¹²)

      dβ = 10  7.077

      dβ = 70. 77 dβ

3 0
3 years ago
An object, whose mass is 0.660 kg, is attached to a spring with a force constant of 132 N/m. The object rests upon a frictionles
Arada [10]

Answer:

See below

Explanation:

a)  Spring force at release =  k * d = 132 N/m * .120 m = 15.84 N

b) F = ma

   15.84 = (.660 kg)(a)     a = 24 m/s^2

c) Toward the left ....the object is accelerated to the left

7 0
1 year ago
A girl with a mass of 27 kg is playing on a swing. There are three main forces
N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

  • Mass of the girl, m = 27 kg
  • Speed of the girl, v = 3 m/s
  • Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0
2 years ago
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