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Bad White [126]
3 years ago
14

You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s

with respect to the ground due north.
1) What is the speed of the plane with respect to the ground?
2) What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east).
3) How far east will the plane travel in 1 hour?
Physics
1 answer:
BlackZzzverrR [31]3 years ago
6 0
1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>
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Three different people weight a standard mass of 2.00 g on the same balance. Each person obtains a reading of 2.32 g for the mas
Lubov Fominskaja [6]

Answer: precise

Explanation:

Three different people weight a standard mass of 2.00 g on the same balance. Each person obtains a reading of 2.32 g for the mass of the standard. These results imply that the balance that was used is precise.

Precision can be defined as the closeness of measured values to each other, for a measuring equipment it is the closeness of the values of readings obtained at different times to each other. It does not necessarily means the measurements are accurate(closeness to the actual value). Therefore, in the case above where three different people measured the same mass on the same balance, and each of them obtained the same value which is different from the standard value. We can say the balance used is precise because the three readings are the same.

5 0
3 years ago
If a ????=87.5 kgm=87.5 kg person were traveling at ????=0.900????v=0.900c , where ????c is the speed of light, what would be th
Diano4ka-milaya [45]

Answer:

\frac{K.E_r}{K.E}=2.875

Explanation:

Given:

mass, m = 87.5kg

Velocity, V = 0.900c

now,

the relativistic kinetic energy id given as:

K.E_r=(\gamma-1)mc^2 ...........(1)

where,

\gamma = relativistic factor, given as; \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

Now, the classical kinetic energy is given as:

K.E = \frac{1}{2}mv^2    ..........(2)

Dividing the equation (1) by (2) we get

\frac{K.E_r}{K.E}=\frac{(\gamma-1)mc^2}{\frac{1}{2}mv^2}

or

\frac{K.E_r}{K.E}=\frac{(\gamma-1)c^2}{\frac{1}{2}v^2}

substituting the values in the equation we get,

\frac{K.E_r}{K.E}=\frac{(\frac{1}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}-1)c^2}{\frac{1}{2}\times(0.90c)^2}

or

\frac{K.E_r}{K.E}=2.875

5 0
3 years ago
An electric motor exerts a constant torque of 10 Nm on a grindstone mounted on its shaft. The moment of inertia of the grindston
Dmitry_Shevchenko [17]

Answer:

Angular acceleration = 5 rad /s ^2

Kinetic energy = 0.391 J

Work done = 0.391 J

P =6.25 W

Explanation:

The torque is given as moment of inertia × angular acceleration

angular acceleration = torque/ moment of inertia

= 10/2= 5 rad/ s^2

The kinetic energy is = 1/2 Iw^2

w = angular acceleration/time

=5/8= 0.625 rad /s

1/2 × 2× 0.625^2

=0.391 J

The work done is equal to the kinetic energy of the motor at this time

W= 0.391 J

The average power is = torque × angular speed

= 10× 0.625

P = 6.25 W

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