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r-ruslan [8.4K]
2 years ago
14

A student took a calibrated 200.0 gram mass, weighed it on a laboratory balance, and found it read 186.5 g. What was the student

’s percent error?
Physics
1 answer:
snow_lady [41]2 years ago
7 0

Answer:

The original or accepted value for the percent by mass of water in a hydrate = 36%

Percen by mass of water in the hydrate determined by the student

in the laboratory = 37.8%

So the difference between the actual and the percent by mass in water determined by student = (37.8 - 36.0)%

            = 1.8%

So the percentage of error made by the student = (1.8/36) * 100 percent

                                                                            = (18/360) * 100 percent

                                                                            = ( 1/20) * 100 percent

                                                                             = 5 percent

So the student makes an error of 5%. Option "1" is the correct option.

Explanation:

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In Fig. 4-41, a ball is thrown up onto a roof, landing 4.00 s later at height h ???? 20.0 m above the release level. The ball’s
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"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) v_{i} = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd  equation of motion

h = v_{y_{f}}t + \frac{1}{2}gt^{2}

-20 = v_{y_{f}} (4) + 0.5(-9.8)(4)²

v_{y_{f}} (4) = 58.4

v_{y_{f}}  = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

v_{f} = v_{y_{f}} / sinθ

v_{f} = 14.6 / sin 60

v_{f} = 16.86 m/s

v_{x_{f}} = v_{f}cosθ

v_{x_{f}} = 16.86 cos 60

v_{x_{f}} = 8.43 m/s

To calculate the horizontal distance

d = v_{y_{f}} t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

v_{x_{f}} = v_{x_{i}}

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = (v_{y_{f}} - v_{y_{i}} ) / t

9.8 =  14.6 - v_{y_{i}}) / 4

v_{y_{i}} = 24.6 m/s

v_{i} = \sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }

v_{i} = \sqrt{8.43^{2}+24.6^{2}}

v_{i} = 26 m/s

c)

cos β = v_{x_{i}} / v_{i}

β = cos⁻¹ (8.43 / 26)

β = 71.08°

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Answer:

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E = 7056 kJ

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