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3 years ago

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A superball and a clay ball are dropped from a height of 10cm above a tabletop. They have the same mass 0.05kg and the same size

. The superball bounces off the table and rises back to the same height. The clay ball sticks to the table. For the superball in the previous question, if it was in contact with the table for 34.3ms, calculate the average force exerted on the ball by the table. Hint: First calculate the momentum before and after hitting the table. Don't forget the gravitational force.

1 answer:

Olenka [21]3 years ago

5 0

Answer:4.08 N

Explanation:

Given data

superball dropped from a height of 10 cm

Mass of ball $\left ( m\right )$ =0.05kg

time of contact $\left ( t\right )$ =34.3 $\times 10^{-3}$ s

Now we know impulse = $Force\times time\ of\ contact$ =Change in momentum

$F_{average}\times t$ = $m\left ( v-(-v)\right )$

and velocity at the bottom is given by

v= $\sqrt{2gh}$

$F_{average}\times 34.3\times 10^{-3}$ = $0.05\left ( 1.4-(-1.4)\right )$

$F_{average}$ =4.081N

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3 years ago

If two stars have the same absolute magnitude, what can be a reason for the difference in their brightness?

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B. their distances from the sun.

Explanation:

Absolute Magnitude:

Astronomers defines the absolute magnitude of a stars brightness in terms of how bright a star appears from a standard distance of 10 parsecs. Parsec is a unit of distance in astronomy. 10 parsecs is equal to 32.6 light years.

Apparent Magnitude:

Apparent magnitude of a star refers to how bright the star appears at its distance from the Earth.

If two stars have the same absolute magnitude but their apparent magnitude differs, the reason is that the distance of both the stars from the Earth varies. Hence their brightness differs when measured from Earth. The farther a star is from the Earth, the fainter its brightness.

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3 years ago

A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140

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Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

$u = 0\\\\v = 2.35\ \frac{m}{sec}\\\\d = 5.0 \ m\\\\$

Using formula:

$v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\$

$= 0.55 \ \frac{m}{sec^2}\\\\$

$F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\$

Calculating the Work by net force

$W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\$

The above work is converted into thermal energy.

Now,

$F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\ by\ friction = -547.64 \ J$

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3 years ago

During an investigation, a scientist heated 123.6 g of copper carbonate till it decomposed to form a black residue. The total ma

zloy xaker [14]

Answer:

See below explanation

Explanation:

The correspondent chemical reaction for copper carbonate decomposed by heat is:

CuCO₃ (s) → CuO (s) + CO₂ (g)

Considering all molar mass (MM) for each element ( we consider rounded numbers) :

MM CuCO₃ = 123 g/mol

MM CuO = 79 g/mol

MM CO₂ = 44 g/mol

Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.

So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which "escaped" from our system, since it is a gas compound (CO₂)

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4 years ago

A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po

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Answer:

Following are the solution to this question:

Explanation:

Law:

$\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}$

$= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}$

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