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AfilCa [17]
3 years ago
10

A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student

’s percent error?
a
9.3
b
6.6
c
6.2
d
16.5
e
7.2
Physics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

B. 6.6%

Explanation:

The percentage error of a measurement can be calculated using the formula;

Percent error = (experimental value - accepted value / accepted value) × 100

In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.

Hence, the percentage error can be calculated thus;

Percent error = (266.5-250.0/250.0) × 100

Percent error = 16.5/250 × 100

Percent error = 0.066 × 100

Percent error = 6.6%

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katen-ka-za [31]

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5 0
2 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
2 years ago
You drop a batitin a stationary elevator and the ball hits the floor in o 50 s. How long does it take for the ball to hit the fl
solmaris [256]

Answer:

option (a) 0.61 s

Explanation:

Given;

Time taken by the ball to reach the ground  = 0.50 s

Let us first calculate the distance through which the ball falls on the ground

from the Newton's equation of motion, we have

s=ut+\frac{1}{2}at^2

where,  

s is the distance

a is the acceleration

t is the time

here it is the case of free fall

thus, a = g = acceleration due to gravity

u =  initial speed of the ball = 0

on substituting the values, we get

s=0\times 0.50+\frac{1}{2}\times9.8\times0.50^2

or

s = 1.225 m

Now,

when the elevator is moving up with speed of 1.0 m/s

the initial speed of the ball = -1.0 m/s   (as the elevator is moving in upward direction)

thus , we have

s=ut+\frac{1}{2}at^2

or

1.225=-1.0\times\ t+\frac{1}{2}\times9.8t^2

or

4.9t^2 - t  - 1.225 = 0

or

t = 0.612 s

hence, the correct answer is option (a) 0.61 s

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Gnesinka [82]

Answer:

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