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Lapatulllka [165]
3 years ago
13

The precision of a laboratory instrument is ± 0.05 g. The accepted value for your measurement is 7.92 g. Which measurements are

in the accepted range? Check all that apply.
a) 7.85 g
b)7.89 g
c) 7.91 g
d) 7.97 g
e) 7.99 g
Physics
1 answer:
Step2247 [10]3 years ago
8 0

Answer:

b)7.89 g  

c) 7.91 g

d) 7.97 g

Explanation:

The accepted value for the measurement is 7.92 g, while the precision of the instrument is 0.05 g. Therefore, the interval of the accepted values will be the central value plus and minus the precision, so:

- Minimum accepted value: 7.92 g - 0.05 g = 7.87 g

- Maximum accepted value: 7.92 g + 0.05 g = 7.97 g

So, the accepted interval is

7.87 g - 7.97 g

Therefore, all the values that lie within this interval are accepted. So the accepted values are:

b)7.89 g  

c) 7.91 g

d) 7.97 g

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3 years ago
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Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

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v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

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Δy = 4.91 m

The distance between the ball and the crossbar is:

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5 0
3 years ago
A sports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. What is its acceleration and how ma
Flauer [41]

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car.  It can be calculated using third equation of motion as :

v^2-u^2=2ad

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{0-(22.22)^2}{2\times 80000}

a=-0.00308\ m/s^2

Value of g, g=9.8\ m/s^2

a=\dfrac{-0.00308}{9.8}\ m/s^2

a=(-0.000314)\ g\ m/s^2

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8 0
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 the focal length <span> is much more decent  for a concave, and also worse</span><span> for a convex mirror. When the image that is given, distance is good and decent, images are always on the same area of the mirror as the object given , and it is not fake.  images distance  is </span>never positive <span>, the image is on the oppisite side of  the mirror, so the image must be virtual.</span>
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tigry1 [53]

Answer:

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