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3 years ago

9

You and a friend each drive 58 km. You travel at 89. km/h, your friend at 94 km/h. How long will your friend wait for you at the

end of the trip?

1 answer:

alexgriva [62]3 years ago

5 0

Answer:

2.1 minutes/ 126 seconds

Explanation:

Distance = speed x time

We can rearrange this equation for time:

Time = Distance/Speed

<u>For you (89km/h):</u>

Distance = 58km, Speed = 89km/h

Therefore time = 58/89 = 0.652 (3dp) hours

<u>For friend:</u>

Distance = 58km, Speed = 94km/h

Therefore time = 58/94 = 0.617 (3dp) hours

Difference in time = 0.652 - 0.617 = 0.035 hours.

Convert to minutes: 0.035 x 60 (because 60 min in hour) = 2.1

Your friend will be waiting for 2.1 minutes, or 126 seconds (2.1 x 60).

Hope this helped!

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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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Answer

given,

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mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

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using conservation of energy

$\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}$

$\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}$

$\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}$

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yaroslaw [1]

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<h3>Resistivity of a material</h3>

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