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ycow [4]
2 years ago
10

In Cynthia's physics class, she must turn in four lab reports. For each report, she earns extra credit if her percent error is l

ess than 5%.5%. For her first lab report, she calculated the value of the acceleration due to gravity three times. She got the following three values. 9.56 ms2 9.56 ms2 9.72ms2 9.72ms2 9.88ms2 9.88ms2 The accepted value of the acceleration due to gravity is 9.8ms2.9.8ms2. Which option is closest to Cynthia's percent error? Remember: Take the average of the three measurements before calculating the percent error.
a. about 8%8%.
b. about 1%1%.
c. about 0.8%0.8%.
d. about 5%.
Physics
1 answer:
MariettaO [177]2 years ago
7 0

Answer:

error = 0.82%

Explanation:

The three readings of her gravity calculation is given as

g_1 = 9.56 m/s^2

g_2 = 9.72 m/s^2

g_3 = 9.88 m/s^2

now the mean value of gravity calculation is given as

g_{mean} = \frac{9.56 + 9.72 + 9.88}{3}

g_{mean} = 9.72 m/s^2

now the percentage error in her calculation is given as

error = \frac{9.8 - 9.72}{9.8} \times 100

error = 0.82%

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Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h
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2 years ago
What is the thickness required of a masonry wall having thermal conductivity 0.75 W/m K if the heat rate is to be 80% of the hea
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Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
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Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

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The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

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The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

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Hence, the deceleration of the car, a = = -4.32 m/s²

5 0
2 years ago
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Answer:

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This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

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The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

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1 year ago
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