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Alekssandra [29.7K]
4 years ago
7

The mass of the Sun is 2 × 1030 kg, and the mass of Saturn is 5.68 × 1026 kg. The distance between Saturn and the Sun is 9.58 AU

. Veronica is solving the following equation to calculate the orbital period of Saturn, but there is an error in the equation. T = What should Veronica change to correct the equation? change the position of 2 x 1030 kg and 9.58 AU change 2 x 1030 kg to 5.68 x 1026 kg change the square root to a cube root change 9.58 AU to the distance in meters
Physics
2 answers:
GREYUIT [131]4 years ago
5 0
Not the answer but the answer will nit be  B. hope yall find the answer
USPshnik [31]4 years ago
5 0

The correct answer is D.

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What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

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MaRussiya [10]
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Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of
ololo11 [35]

Explanation:

It is given that,

The separation between two parallel wires, r = 5.6 cm = 0.056 m

Current in both the wires is 2.65 A

(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :

\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m

(b) As the current is in same direction, the wires will attract each other.

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Explanation:

pls brainlist too

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