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aivan3 [116]
4 years ago
15

The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t

o within 0.1kmh. What is the potential error in her calculated kinetic energy, as a result of the imprecision in the measurement of her velocity?
Physics
1 answer:
jek_recluse [69]4 years ago
4 0

Answer:

e=3367.2J

%e=1.43%

Explanation:

From the exercise we know two information. The real speed and the experimental measured by the speedometer

v_{r}=10km/h=2.77m/s

Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

%e=\frac{E_{r}-E_{e}}{E_{r}}x100=\frac{(234023-230656)J}{234023J}x100=1.43%

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The density of sample is 5 g/cm3

Given:

volume of sample = 20 cm3

mass of sample = 100 grams

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density = mass/volume

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d = 5 g/cm3

So, density of sample is 5 g/cm3

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2 years ago
A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.
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<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

           Acceleration, a = -9.81 m/s²

            Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

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                t² - 26.28 t = 0

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After 26.28 seconds projectile returns 26.28 seconds.

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Explanation:

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Calculate the induced electric field (in V/m) in a 40-turn coil with a diameter of 18 cm that is placed in a spatially uniform m
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Answer:

ε = 6.617 V

Explanation:

We are given;

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The formula for the induced electric field(E.M.F) is given by;

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A is area

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While N,B and t remain as earlier described.

Area = π(d²/4) = π(0.18²/4) = 0.02545

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