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Oduvanchick [21]
2 years ago
10

A glass ball of radius 3.74 cm sits at the bottom of a container of milk that has a density of 1.04 g/cm3. The normal force on t

he ball from the container's lower surface has magnitude 9.03 x 10-2 N. What is the mass of the ball?
Physics
1 answer:
Gelneren [198K]2 years ago
5 0

Answer:

The mass of the ball is 0.23 kg

Explanation:

Given that

radius ,r= 3.74 cm

Density of the milk ,ρ = 1.04 g/cm³ = 1.04  x 10⁻³ kg/cm³

Normal force ,N= 9.03 x 10⁻² N

The volume of the ball V

V=\dfrac{4}{3}\pi r^3

V=\dfrac{4}{3}\times \pi \times 3.74^3\ cm^3

V= 219.13 cm³

The bouncy force on the ball = Fb

Fb = ρ V g

Fb  + N = m g

m=Mass of the ball = Density x volume

m = γ V    , γ =Density of the Ball

ρ V g  + N =  γ V g               ( take g= 10 m/s²)

\gamma =\dfrac{N+\rho V g}{V g}

\gamma =\dfrac{9.03\times 10^{-2}+1.04\times 10^{-3}\times 219.13\times  10}{219.13\times 10}

γ = 0.00108 kg/cm³

m = γ V

m = 0.00108 x 219.13

m= 0.23 kg

The mass of the ball is 0.23 kg

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2 years ago
Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5,
finlep [7]

Answer:

correct answer is 1 and 3

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In direct measurement with an instrument, the precision or absolute error of the instrument is given by its appreciation, in this case we see that the measurements have two decimal places, so the appreciation of the instrument must be 0.01 cm

Based on this appreciation, the valid measurements are 5.52 and 5.5.

the other two measurements have errors much higher than the assessment of the instrument, for which there must have been some errors in the measurement.

The correct answer is 1 and 3

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3 years ago
An ice rink is 200 feet long and 85 feet wide. The four corners of the rink are rounded to a radius of 24 feet. What is the surf
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Answer:

16,506 ft²

Explanation:

There are different ways you can divide the area using rectangles and circles.  One way is to find the area of the entire width and length, then subtract the empty areas in the corners.

If we take the empty areas and put them together, we find their area is the area of a square minus the area of a circle.

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A = WL − (4 − π) r²

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5 0
2 years ago
An object is (starting at rest) begins to plummet towards the earth. Assuming it is in a vacuum, how fast is the object travelin
Eva8 [605]

Answer:

–50.96

Explanation:

The following data were obtained from the question:

Initial velocity (Vᵢ) = 0 m/s

Acceleration (a) = – 9.8 m/s²

Time (t) = 5.2 s

Final velocity (Vբ) =.?

Acceleration is simply defined as the change of velocity with time. Mathematically, it is expressed as:

Acceleration (a) = Final velocity (Vբ) – Initial velocity (Vᵢ) /Time (t) =

a = (Vբ – Vᵢ) / t

With the above formula, we can determine how fast the object is traveling after 5 s as follow:

Initial velocity (Vᵢ) = 0 m/s

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Time (t) = 5.2 s

Final velocity (Vբ) =.?

a = (Vբ – Vᵢ) / t

– 9.8 = (Vբ – 0) / 5.2

– 9.8 = Vբ / 5.2

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Vբ = –9.8 × 5.2

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Therefore, the object is traveling at

–50.96 m/s

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