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Mandarinka
3 months ago
12

If one of the satellites is at a distance of 20,000 km from you, what percent accuracy in the distance is required if we desire

a 2-meter uncertainty
Physics
1 answer:
Maru [3.3K]3 months ago
8 0
<span>To express uncertainty as a percentage, divide the uncertainty by the measured value and multiply by 100%.

(2 m / 20,000,000 m) × 100% = 0.00001%

This means the percent uncertainty is 0.00001%.

Subtracting this from 100% gives the percent accuracy:
Percent accuracy = 100% - 0.00001% = 99.99999%

Therefore, the percent accuracy required is 99.99999%.</span>
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Two trains are headed towards each other on the same track unbeknownst to the engineers. One departs San Francisco. Its average
ValentinkaMS [3465]

Answer:

7.166 hours = 430 minutes.

Explanation:

As both trains are approaching each other on the same track, their relative speed is the sum of their individual speeds. Hence, the time until they intersect (and inevitably collide) is determined by how long it takes for speeds of 65 mph and 55 mph to cover the total distance of 860 miles. One train will cover part of the distance, while the other will cover the remainder. To calculate the required time, we can apply the formula:

1 hour ---> 120 miles

X ----> 860 miles; hence X = (860 miles * 1 hour)/120 miles = 43/6 hours = 7.16666 hours. To convert this into minutes, recall that 1 hour equals 60 minutes; therefore, 43/6 hours * 60 minutes/hour = 430 minutes.

7 0
2 months ago
A spinning wheel is slowed down by a brake, giving it a constant angular acceleration of 25.60 rad/s2. During a 4.20-s time inte
Yuliya22 [3333]

We will use the equations of rotational kinematics,

\theta =\theta _{0} + \omega_{0} t+ \frac{1}{2}\alpha t^2             (A)

\omega^2= \omega^2_{0} +2\alpha\theta                                     (B)                                          

Here, \theta and \theta _{0} denote the final and initial angular displacements, respectively, whereas \omega and \omega_{0} represent final and initial angular velocities, and \alpha is the angular acceleration.

We are provided with \alpha = - 25.60 \ rad/s^2, \theta = 62.4 \ rad and t = 4.20 \ s.

By substituting these values into equation (A), we have

62.4 \ rad = 0 + \omega_{0} 4.20 \ s + \frac{1}{2} (- 25.60 \ rad/s^2) ( 4.20)^2 \\\\ \omega_{0} = \frac{220.5+ 62.4 }{4.20} =67.4 \ rad/s

Now, using equation (B),

\omega^2=(67.4 \ rad/s)^2 + 2 (- 25.60 \ rad/s^2)62.4 \ rad \\\\\ \omega = 36.7 \ rad/s

This indicates that the wheel's angular speed at the 4.20-second mark is 36.7 rad/s.

4 0
2 months ago
a force of 6lbs acts on an object with a weight of 35 lbs on earth. determine the objects acceleration. final answer must be 5.5
kicyunya [3294]
Weight of the object = 35 lbs
F = ma
m = F/a = 35/32 (with acceleration of 32 ft/s²)
m= 1.09

Again applying the same formula,
a = F/m
a= 6/1.09
a= 5.489

Thus, the acceleration is approximately 5.5 ft/s²!!
5 0
2 months ago
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Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and com
Ostrovityanka [3204]

Answer:

The book is titled Solid State or Condensed Matter

Explanation:

8 0
2 months ago
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